1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef pair<int, int> pii;
 4 typedef long long ll;
 5 #define ff first
 6 #define ss second
 7 
 8 int a[1000010];
 9 int main() {
10     int n, k; scanf("%d%d", &n, &k);
11     for (int i = 1; i <= n; i++)a[i] = 1;
12     int ans = 0, cur = 0;
13     while (k--) {
14         int m, cnt = cur; scanf("%d", &m);
15         for (int i = 1; i * m <= n; i++) {
16             if (!a[i * m])cnt--;
17             else cnt++;
18             a[i * m] ^= 1;
19         }
20         ans = max(ans, cnt);
21         //cout<<cnt<<endl;
22         cur = cnt;
23     }
24     printf("%d\n", ans);
25 }

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef pair<int, int> pii;
 4 typedef long long ll;
 5 #define ff first
 6 #define ss second
 7 
 8 int dl[20][100010], dr[20][100010];
 9 int query(int d[][100010], int l, int r, int maxx) {
10     int t = log2(r - l + 1);
11     if (maxx)return max(d[t][l], d[t][r - (1 << t) + 1]);
12     return min(d[t][l], d[t][r - (1 << t) + 1]);
13 }
14 void init(int d[][100010], int a[], int len, int maxx) {
15     for (int i = 0; i < len; i++)d[0][i] = a[i];
16     int t = 1;
17     for (int i = 1; t <= len; i++) {
18         for (int j = 0; j + t < len; j++)
19             if (maxx)d[i][j] = max(d[i - 1][j], d[i - 1][j + t]);
20             else d[i][j] = min(d[i - 1][j], d[i - 1][j + t]);
21         t <<= 1;
22     }
23 }
24 
25 int L[100010], R[100010];
26 int main() {
27     int n; scanf("%d", &n);
28     for (int i = 0; i < n; i++)scanf("%d%d", &L[i], &R[i]);
29     init(dl, L, n, 1);
30     init(dr, R, n, 0);
31 
32     int ans = 1;
33     for (int i = 1; i <= n; i++) {
34         int l = ans, r = n - i + 1;
35         while (l <= r) {
36             int mid = l + r >> 1;
37             //cout<<i<<" "<<mid<<" "<<query(dl,i-1,i+mid-2,1)<<" "<<query(dr,i-1,i+mid-2,0)<<"\n";
38             if (query(dr, i - 1, i + mid - 2, 0) - query(dl, i - 1, i + mid - 2, 1) + 1 >= mid) {
39                 ans = max(ans, mid); l = mid + 1;
40             } else r = mid - 1;
41         }
42     }
43     printf("%d\n", ans);
44 }

 1 /* basic header */
 2 #include <bits/stdc++.h>
 3 /* define */
 4 #define ll long long
 5 #define dou double
 6 #define pb emplace_back
 7 #define mp make_pair
 8 #define sot(a,b) sort(a+1,a+1+b)
 9 #define rep1(i,a,b) for(int i=a;i<=b;++i)
10 #define rep0(i,a,b) for(int i=a;i<b;++i)
11 #define eps 1e-8
12 #define int_inf 0x3f3f3f3f
13 #define ll_inf 0x7f7f7f7f7f7f7f7f
14 #define lson (curpos<<1)
15 #define rson (curpos<<1|1)
16 /* namespace */
17 using namespace std;
18 /* header end */
19 
20 const int maxn = 200010, lim = 100001;
21 struct Point {
22     int x, y;
23 } p[maxn];
24 int n, x, y, maxx[maxn], minn[maxn], monument[maxn];
25 
26 void init() {
27     for (int i = 0; i < maxn; i++) {
28         p[i].x = p[i].y = monument[i] = 0;
29         maxx[i] = -1;
30     }
31     memset(minn, 0x3f, sizeof(minn));
32 }
33 
34 int main() {
35     while (~scanf("%d%d", &x, &y)) {
36         init();
37         scanf("%d", &n);
38         for (int i = 0; i < n; i++) {
39             scanf("%d%d", &p[i].x, &p[i].y);
40             maxx[p[i].x] = max(maxx[p[i].x], p[i].y);
41             minn[p[i].x] = min(minn[p[i].x], p[i].y);
42         }
43         int cnt = 0;
44         for (int i = 0; i < lim; i++) {
45             if (maxx[i] != -1 && minn[i] != 0x3f3f3f3f) {
46                 monument[cnt++] = maxx[i];
47                 monument[cnt++] = minn[i];
48             }
49         }
50         sort(monument, monument + cnt);
51         ll ans = x - 1, mid = monument[(cnt - 1) / 2];
52         for (int i = 0; i < lim; i++)
53             if (maxx[i] != -1 && minn[i] != 0x3f3f3f3f)
54                 ans += (ll)maxx[i] - minn[i] + abs(mid - maxx[i]) + abs(mid - minn[i]);
55         printf("%lld\n", ans);
56     }
57     return 0;
58 }

 1 #include <iostream>
 2 #include <string>
 3 using namespace std;
 4 char str[50001];
 5 string names[50001];
 6 int d[50001];
 7 int validate(int x) {
 8     if (x > 49) return 49;
 9     if (x < -50) return -50;
10     return x;
11 }
12 int main() {
13     int n;
14     int sum = 0;
15     scanf("%d", &n);
16     for (int i = 0; i < n; ++i) {
17         scanf("%s%d", str, &d[i]);
18         d[i] *= 100;
19         sum += d[i];
20         names[i] = str;
21     }
22     int l = 10000 - sum - 49 * (n - 1);
23     int r = 10000 - sum + 50 * (n - 1);
24     if (r > 49) r = 49;
25     if (l < -50) l = -50;
26     if (l <= r) {
27         for (int i = 0; i < n; ++i) {
28             printf("%s %.2f %.2f\n", names[i].c_str(), max(0, d[i] + l) / 100.0, min(10000, d[i] + r) / 100.0);
29         }
30     } else printf("IMPOSSIBLE\n");
31 }

  1 /* basic header */
  2 #include <bits/stdc++.h>
  3 /* define */
  4 #define ll long long
  5 #define dou double
  6 #define pb emplace_back
  7 #define mp make_pair
  8 #define sot(a,b) sort(a+1,a+1+b)
  9 #define eps 1e-8
 10 #define int_inf 0x3f3f3f3f
 11 #define ll_inf 0x7f7f7f7f7f7f7f7f
 12 #define lson (curpos<<1)
 13 #define rson (curpos<<1|1)
 14 /* namespace */
 15 using namespace std;
 16 /* header end */
 17 using namespace std;
 18 
 19 const int maxn = 1e5 + 10;
 20 struct Node {
 21     int x, y, z, id, cnt;
 22     bool operator<(const Node &rhs)const {
 23         return x != rhs.x ? x < rhs.x : y != rhs.y ? y < rhs.y : z < rhs.z;
 24     }
 25 } info[maxn], temp[maxn];
 26 int n, m, num = 0, sum = 0, a[maxn], dis1[maxn], dis2[maxn], dis3[maxn], vis[maxn], head[maxn];
 27 vector<pair<int, int>>g[maxn];
 28 priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>>q;
 29 
 30 int lowbit(int x) {
 31     return x & -x;
 32 }
 33 
 34 void add(int pos, int val) {
 35     while (pos < maxn) a[pos] += val, pos += lowbit(pos);
 36 }
 37 
 38 int query(int pos) {
 39     int ret = 0;
 40     while (pos) ret += a[pos], pos -= lowbit(pos);
 41     return ret;
 42 }
 43 
 44 void dijkstra(int *dis, int st) {
 45     for (int i = 0; i <= n; ++i) dis[i] = int_inf;
 46     dis[st] = 0;
 47     while (q.size()) q.pop();
 48     q.push(mp(dis[st], st));
 49     while (q.size()) {
 50         pair<int, int> temp = q.top(); q.pop();
 51         if (temp.first > dis[temp.second]) continue;
 52         for (int i = 0; i < (int)g[temp.second].size(); ++i) {
 53             if (dis[temp.second] + g[temp.second][i].first < dis[g[temp.second][i].second]) {
 54                 dis[g[temp.second][i].second] = dis[temp.second] + g[temp.second][i].first;
 55                 q.push(mp(dis[g[temp.second][i].second], g[temp.second][i].second));
 56             }
 57         }
 58     }
 59 }
 60 
 61 void init() {
 62     for (int i = 0; i < maxn; i++) vis[i] = 0;
 63 }
 64 
 65 void cdq(int l, int r) {
 66     if (r - l <= 1) return;
 67     int mid = (r + l) >> 1;
 68     cdq(l, mid); cdq(mid, r);
 69     int lPtr = l, rPtr = mid, cnt = 0;
 70     while (lPtr < mid && rPtr < r) {
 71         if (info[lPtr].y <= info[rPtr].y) {
 72             add(info[lPtr].z, 1);
 73             temp[cnt++] = info[lPtr++];
 74         } else {
 75             int curr = query(info[rPtr].z);
 76             if (curr) vis[info[rPtr].id] = info[rPtr].cnt;
 77             temp[cnt++] = info[rPtr++];
 78         }
 79     }
 80     while (rPtr < r) {
 81         int curr = query(info[rPtr].z);
 82         if (curr) vis[info[rPtr].id] = info[rPtr].cnt;
 83         temp[cnt++] = info[rPtr++];
 84     }
 85     for (int i = l; i < lPtr; ++i) add(info[i].z, -1);
 86     while (lPtr < mid) temp[cnt++] = info[lPtr++];
 87     for (int i = 0; i < cnt; ++i) info[i + l] = temp[i];
 88 }
 89 
 90 int main() {
 91     scanf("%d%d", &n, &m); sum = n, num = 0;
 92     while (m--) {
 93         int u, v, w;
 94         scanf("%d%d%d", &u, &v, &w);
 95         g[u].pb(mp(w, v)); g[v].pb(mp(w, u));
 96     }
 97     // maintain distance
 98     dijkstra(dis1, 0); dijkstra(dis2, 1); dijkstra(dis3, 2);
 99     for (int i = 0; i < n; ++i)  info[i].x = dis1[i], info[i].y = dis2[i], info[i].z = dis3[i], head[++num] = info[i].z;
100     sort(head + 1, head + 1 + num);
101     // li san hua
102     num = unique(head + 1, head + 1 + num) - head - 1;
103     for (int i = 0; i < n; ++i) info[i].z = lower_bound(head + 1, head + 1 + num, info[i].z) - head;
104     sort(info, info + n);
105     int p = 1, tot = 0;
106     // spj equal
107     for (int i = 0; i < n; ++i) {
108         if (info[i].x == info[i + 1].x && info[i].y == info[i + 1].y && info[i].z == info[i + 1].z) {
109             p++; continue;
110         }
111         info[i].cnt = p; info[tot++] = info[i]; p = 1;
112     }
113     n = tot;
114     for (int i = 0; i < n; ++i) info[i].id = i;
115     cdq(0, n);
116     for (int i = 0; i < n; ++i) sum -= vis[i];
117     printf("%d\n", sum);
118 }

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 typedef pair<int, int> pii;
  4 typedef long long ll;
  5 #define ff first
  6 #define ss second
  7 
  8 char s[1010][1010];
  9 int vis[1010][1010], dd[8][2] = {-1, 0, -1, -1, -1, 1, 0, 1, 0, -1, 1, 0, 1, -1, 1, 1};
 10 
 11 int abc[3][15] = {{1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1},
 12     {1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1},
 13     {1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1}
 14 };
 15 int add1[20], add2[20];
 16 
 17 int check(int x, int y, int n, int m, char cur) {
 18     for (int i = x; i < x + n; i++)
 19         for (int j = y; j < y + m; j++)
 20             if (s[i][j] != cur || vis[i][j])return 1;
 21     return 0;
 22 }
 23 
 24 int find(int stx, int sty, int x, int y, int cur) {
 25     add1[0] = 0, add1[1] = x, add1[2] = x + y;
 26     for (int i = 3; i < 15; i++)add1[i] = add1[i % 3];
 27     add2[0] = 0, add2[3] = x, add2[6] = x + y, add2[9] = 2 * x + y, add2[12] = 2 * x + 2 * y;
 28     for (int i = 0; i < 5; i++)
 29         for (int j = 1; j < 3; j++)add2[3 * i + j] = add2[3 * i + j - 1];
 30     //for(int i=0;i<15;i++)printf("%d ",add2[i]);printf("\n");
 31 
 32     for (int i = 0; i < 15; i++) {
 33         int addn = 0, addm = 0;
 34         if (i / 3 % 2 == 0)addn = x; else addn = y;
 35         if (i % 3 % 2 == 0)addm = x; else addm = y;
 36 
 37         char findchar = ''.'';
 38         if (abc[cur][i])findchar = ''#'';
 39         if (check(stx + add2[i], sty + add1[i], addn, addm, findchar)) {
 40             //printf("%d\n",i);
 41             //printf("%d %d %d %d\n",stx+add2[i],sty+add1[i],addn,addm);
 42             return 0;
 43         }
 44     }
 45     return 1;
 46 }
 47 
 48 int main() {
 49     int ansa = 0, ansb = 0, ansc = 0;
 50     int n, m; scanf("%d%d", &m, &n);
 51     for (int i = 1; i <= n; i++)scanf("%s", s[i] + 1);
 52     for (int i = 2; i <= n - 1; i++)
 53         for (int j = 2; j <= m - 1; j++)
 54             if (s[i][j] == ''#'') {
 55                 int gg = 0;
 56                 for (int k = 0; k < 8; k++)
 57                     if (s[i + dd[k][0]][j + dd[k][1]] == ''#'') {
 58                         gg++;
 59                         break;
 60                     }
 61                 if (!gg)s[i][j] = ''.'';
 62             }
 63 
 64     for (int i = 2; i <= n - 1; i++)
 65         for (int j = 2; j <= m - 2; j++)
 66             if (s[i][j] == ''.'' && s[i][j + 1] == ''#'' && !vis[i][j]) {
 67                 int a = 1, b = 1;
 68                 for (int k = j + 2; k <= m - 2; k++)
 69                     if (s[i][k] == ''#'')a++;
 70                     else break;
 71                 for (int k = i + 1; k <= n - 2; k++)
 72                     if (s[k][j + 1] == ''#'')b++;
 73                     else break;
 74 
 75                 int x = 2 * a - b, y = 2 * b - 3 * a;
 76                 if (x <= 0 || y <= 0)continue;
 77 
 78                 int gg = 0;
 79                 for (int k = j; k <= j + a + 1; k++)
 80                     if (s[i - 1][k] == ''#'' || s[i + b][k] == ''#'') {
 81                         gg++;
 82                         break;
 83                     }
 84                 if (gg)continue;
 85                 for (int k = i; k <= i + b - 1; k++)
 86                     if (s[k][j] == ''#'' || s[k][j + a + 1] == ''#'') {
 87                         gg++;
 88                         break;
 89                     }
 90                 if (gg)continue;
 91 
 92                 int cur = 0;
 93                 if (s[i][j + x + 1] == ''#'' && s[i + x + y][j + x + 1] == ''#'' && s[i + 2 * (x + y)][j + x + 1] == ''.'')cur = 1;
 94                 if (s[i][j + x + 1] == ''#'' && s[i + x + y][j + x + 1] == ''#'' && s[i + 2 * (x + y)][j + x + 1] == ''#'')cur = 2;
 95                 if (s[i][j + x + 1] == ''#'' && s[i + x + y][j + x + 1] == ''.'' && s[i + 2 * (x + y)][j + x + 1] == ''#'')cur = 3;
 96 
 97                 if (!cur)continue;
 98 
 99                 //printf("%d %d %d\n",i,j,cur);
100 
101                 if (find(i, j + 1, x, y, cur - 1)) {
102                     for (int visx = i - 1; visx <= i + b; visx++)
103                         for (int visy = j; visy <= j + a + 1; visy++)
104                             vis[visx][visy]++;
105                     if (cur == 1)ansa++;
106                     else if (cur == 2)ansb++;
107                     else ansc++;
108                 }
109 
110             }
111     /*for(int i=1;i<=n;i++,printf("\n"))
112         for(int j=1;j<=m;j++)printf("%d",vis[i][j]);*/
113     printf("%d %d %d\n", ansa, ansb, ansc);
114 }

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 char str[1001];
 4 int dp[1001][1001];
 5 long long strhash[1001];
 6 long long M = 1000000007;
 7 long long p[1001];
 8 long long getHash(int l, int r) {
 9     l++; r++;
10     return (strhash[r] - (strhash[l - 1] * p[r - l + 1]) % M + M) % M;
11 }
12 int getKey(char ch) {
13     if (ch >= ''a'' && ch <= ''z'') return ch - ''a'' + 1;
14     if (ch >= ''A'' && ch <= ''Z'') return ch - ''A'' + 27;
15     return ch - ''0'' + 54;
16 }
17 int solve(int l, int r) {
18     if (~dp[l][r]) return dp[l][r];
19     if (l == r) {
20         dp[l][r] == 1;
21         return 1;
22     }
23     int ans = 99999999;
24     for (int i = 1; i <= r - l; ++i) {
25         int j;
26         long long rep = getHash(l, l + i - 1);
27         for (j = l; j <= r - i + 1; j += i) if (rep != getHash(j, j + i - 1)) break;
28         int f1, f2;
29         f1 = f2 = 0;
30         f1 = solve(l, l + i - 1);
31         if (j <= r) f2 = solve(j, r);
32         ans = min(ans, f1 + f2);
33     }
34     dp[l][r] = ans;
35     return ans;
36 }
37 int main() {
38     int len;
39     scanf("%d", &len);
40     scanf("%s", str);
41     p[0] = 1;
42     for (int i = 1; i <= len; ++i) p[i] = p[i - 1] * 71 % M;
43     strhash[0] = 0;
44     for (int i = 1; i <= len; ++i) strhash[i] = ((strhash[i - 1] * 71) % M + getKey(str[i - 1])) % M;
45     //cout<<getHash(0,10)<<'' ''<<getHash(11,21)<<endl;
46     memset(dp, -1, sizeof(dp));
47     printf("%d\n", solve(0, len - 1));
48 }

#include <bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int n,a[N];
int la()
{
    int pos=0;
    for(int i=1;i<n;i++)
        if(a[pos]<a[i])pos=i;
    if(pos==0||pos==n-1)return 0;
    for(int i=1;i<=pos;i++)
        if(a[i-1]>=a[i])return 0;
    for(int i=pos+1;i<n;i++)
        if(a[i-1]<=a[i])return 0;
    return 1;
}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        cin>>n;
        for(int i=0;i<n;i++)cin>>a[i];
        cout<<(la()?"Yes":"No")<<"\n";
    }
    return 0;
}

It''s Karaoke time! DreamGrid is performing the song Powder Snow in the game King of Karaoke. The song performed by DreamGrid can be considered as an integer sequence , and the standard version of the song can be considered as another integer sequence . The score is the number of integers  satisfying  and .

As a good tuner, DreamGrid can choose an integer  (can be positive, 0, or negative) as his tune and add  to every element in . Can you help him maximize his score by choosing a proper tune?

Input

There are multiple test cases. The first line of the input contains an integer  (about 100), indicating the number of test cases. For each test case:

The first line contains one integer  (), indicating the length of the sequences  and .

The second line contains  integers  (), indicating the song performed by DreamGrid.

The third line contains  integers  (), indicating the standard version of the song.

It''s guaranteed that at most 5 test cases have .

Output

For each test case output one line containing one integer, indicating the maximum possible score.

Sample Input

2
4
1 2 3 4
2 3 4 6
5
-5 -4 -3 -2 -1
5 4 3 2 1

Sample Output

3
1

Hint

For the first sample test case, DreamGrid can choose  and changes  to .

For the second sample test case, no matter which  DreamGrid chooses, he can only get at most 1 match.

 和上次的题目一样，直接做差就好了

#include <bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int a[N];
unordered_map<int,int>M;
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        M.clear();
        int n,ma=0;
        cin>>n;
        for(int i=0;i<n;i++)cin>>a[i];
        for(int i=0,x;i<n;i++)
            cin>>x,M[a[i]-x]++;
        for(auto X:M)ma=max(ma,X.second);
        cout<<ma<<"\n";
    }
    return 0;
}

Now Loading!!!

---

Time Limit: 1 Second       Memory Limit: 131072 KB

---

DreamGrid has  integers . DreamGrid also has  queries, and each time he would like to know the value of

 

for a given number  , where , .

 

Input

There are multiple test cases. The first line of input is an integer  indicating the number of test cases. For each test case:

The first line contains two integers  and  () -- the number of integers and the number of queries.

The second line contains  integers  ().

The third line contains  integers  ().

It is guaranteed that neither the sum of all  nor the sum of all  exceeds .

Output

For each test case, output an integer , where  is the answer for the -th query.

Sample Input

2
3 2
100 1000 10000
100 10
4 5
2323 223 12312 3
1232 324 2 3 5

Sample Output

11366
45619

 有一个向下取整还有向上取整，所以去枚举分母做前缀和优化，卡MLE是真的骚

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=5e5+5,MD=1e9;
ll sum[32][N];
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int n,m;
        cin>>n>>m;
        for(int i=1;i<=n;i++)
            cin>>sum[0][i];
        sort(sum[0]+1,sum[0]+n+1);
        for(int i=1;i<32;i++)
            for(int j=1;j<=n;j++)
            sum[i][j]=sum[0][j]/i+sum[i][j-1];
        ll ans=0;
        for(int i=0,p;i<m;i++)
        {
            cin>>p;
            ll ans1=1,ans2=0;
            for(int j=1;;j++)
            {
                int pos1=lower_bound(sum[0]+1,sum[0]+1+n,ans1+1)-sum[0],pos2=upper_bound(sum[0]+1,sum[0]+1+n,ans1*p)-sum[0]-1;
                if(pos1==n+1)break;
                if(pos1<=pos2)ans2=(ans2+sum[j][pos2]-sum[j][pos1-1])%MD;
                ans1=ans1*p;
            }
            ans=(ans+ans2*(i+1))%MD;
        }
        cout<<ans<<"\n";
    }
    return 0;
}

 I

Magic Points

---

Time Limit: 1 Second       Memory Limit: 65536 KB       Special Judge

---

Given an integer , we say a point  on a 2D plane is a magic point, if and only if both  and  are integers, and exactly one of the following conditions is satisfied:

•  and ;

•  and ;

•  and ;

•  and .

It''s easy to discover that there are  magic points in total. These magic points are numbered from  to  in counter-clockwise order starting from .

DreamGrid can create  magic lines from these magic points. Each magic line passes through exactly two magic points but cannot be parallel to the line  or  (that is to say, the coordinate axes).

The intersections of the magic lines are called dream points, and for some reason, DreamGrid would like to make as many dream points as possible. Can you tell him how to create these magic lines?

Input

There are multiple test cases. The first line of input contains an integer  (about 100), indicating the number of test cases. For each test case, there is only one integer  ().

Output

For each case output  integers  in one line separated by one space, indicating that in your answer, point  and point  is connected by a line for all .

If there are multiple answers, you can print any of them.

Sample Input

3
2
3
4

Sample Output

0 2 1 3
1 4 2 5 3 6
0 6 1 9 3 8 4 10

Hint

The sample test cases are shown as follow:

4n-4个点让你构造直线而且相交点最多，先考虑让他们同方向的，也就是斜率相同的，但是不一定可以构造出来

但是首先要明确的就是按照边走，i连上n+i

最后一条线刚开始直接把对角线的点相连。但是这样有可能会交于同一点，所以是要找最后一点相连还有他的斜率

n==5也是特殊的，太菜了，哎。不过比赛时这个题目还是可以开的

#include <bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int ans[N];
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int n;
        cin>>n;
        if(n==2)cout<<"0 2 1 3\n";
        else if(n==3)cout<<"1 4 2 5 3 6\n";
        else if(n==4)cout<<"0 6 1 9 3 8 4 10\n";
        else if(n==5)cout<<"0 5 1 6 2 7 3 8 11 13\n";
        else
        {
            for(int i=0;i<n-1;i++)cout<<i<<" "<<n+i<<" ";
            cout<<n*4-5<<" "<<n+n<<"\n";
        }
    }
    return 0;
}

J

CONTINUE...?

---

Time Limit: 1 Second       Memory Limit: 65536 KB       Special Judge

---

DreamGrid has  classmates numbered from  to . Some of them are boys and the others are girls. Each classmate has some gems, and more specifically, the -th classmate has  gems.

DreamGrid would like to divide the classmates into four groups , ,  and  such that:

• Each classmate belongs to exactly one group.

• Both  and  consist only of girls. Both  and  consist only of boys.

• The total number of gems in  and  is equal to the total number of gems in  and .

Your task is to help DreamGrid group his classmates so that the above conditions are satisfied. Note that you are allowed to leave some groups empty.

Input

There are multiple test cases. The first line of input is an integer  indicating the number of test cases. For each test case:

The first line contains an integer  () -- the number of classmates.

The second line contains a string  () consisting of 0 and 1. Let  be the -th character in the string . If , the -th classmate is a boy; If , the -th classmate is a girl.

It is guaranteed that the sum of all  does not exceed .

Output

For each test case, output a string consists only of {1, 2, 3, 4}. The -th character in the string denotes the group which the -th classmate belongs to. If there are multiple valid answers, you can print any of them; If there is no valid answer, output "-1" (without quotes) instead.

Sample Input

5
1
1
2
10
3
101
4
0000
7
1101001

Sample Output

-1
-1
314
1221
3413214

---

这个就是凑数的题目，可以凑出来的，贪心选择或者找其中的一段区间都是可以的

#include <bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int ans[N];
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int n;
        cin>>n;
        string s;
        cin>>s;
        long long sum=n*1LL*(n+1)/2;
        if(sum%2)cout<<"-1\n";
        else
        {
            sum/=2;
            for(int i=n-1;i>=0;i--)
            {
                if(sum>i)
                    sum-=(i+1),ans[i]=(s[i]==''1'')?3:1;
                else ans[i]=(s[i]==''1'')?4:2;
            }
            for(int i=0;i<n;i++)cout<<ans[i];
            cout<<"\n";
        }
    }
    return 0;
}

L

Doki Doki Literature Club

---

Time Limit: 1 Second       Memory Limit: 65536 KB

---

Doki Doki Literature Club! is a visual novel developed by Team Salvato. The protagonist is invited by his childhood friend, Sayori, to join their high school''s literature club. The protagonist then meets the other members of the club: Natsuki, Yuri, and the club president Monika. The protagonist starts to participate in the club''s activities such as writing and sharing poetry, and grows close to the four girls. What a lovely story!

A very important feature of the game is its poetry writing mechanism. The player is given a list of various words to select from that will make up his poem. Each girl in the Literature Club has different word preferences, and will be very happy if the player''s poem is full of her favorite words.

The poem writing mini-game (from wikipedia)

BaoBao is a big fan of the game and likes Sayori the most, so he decides to write a poem to please Sayori. A poem of  words  is nothing more than a sequence of  strings, and the happiness of Sayori after reading the poem is calculated by the formula

 

where   is the happiness and  is Sayori''s preference to the word .

 

Given a list of  words and Sayori''s preference to each word, please help BaoBao select  words from the list and finish the poem with these  words to maximize the happiness of Sayori.

Please note that each word can be used at most once!

Input

There are multiple test cases. The first line of input contains an integer  (about 100), indicating the number of test cases. For each test case:

The first line contains two integers  and  (), indicating the number of words and the length of the poem.

For the following  lines, the -th line contains a string consisting of lowercased English letters  () and an integer  (), indicating the -th word and Sayori''s preference to this word. It''s guaranteed that  for all .

Output

For each test case output one line containing an integer  and  strings  separated by one space, indicating the maximum possible happiness and the corresponding poem. If there are multiple poems which can achieve the maximum happiness, print the lexicographically smallest one.

Please, DO NOT output extra spaces at the end of each line, or your answer may be considered incorrect!

A sequence of  strings  is lexicographically smaller than another sequence of  strings , if there exists a  () such that  for all  and  is lexicographically smaller than .

A string  is lexicographically smaller than another string , if there exists a  () such that  for all  and , or  for all  and .

Sample Input

4
10 8
hello 0
world 0
behind 0
far 1
be 2
spring 10
can 15
comes 20
winter 25
if 200
5 5
collegiate 0
programming -5
zhejiang 10
provincial 5
contest -45
3 2
bcda 1
bcd 1
bbbbb 1
3 2
a 1
aa 1
aaa 1

Sample Output

2018 if winter comes can spring be far behind
15 zhejiang provincial collegiate programming contest
3 bbbbb bcd
3 a aa

 仅仅是个sort，读懂题意就可以做了

#include <bits/stdc++.h>
using namespace std;
const int N=1e5+5;
struct T
{
    string s;
    int va;
}a[105];
int cmp(T a,T b)
{
    return a.va>b.va||a.va==b.va&&a.s<b.s;
}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int n,m;
        cin>>n>>m;
        for(int i=0;i<n;i++)cin>>a[i].s>>a[i].va;
        sort(a,a+n,cmp);
        long long sum=0;
        for(int i=0;i<m;i++)
            sum+=(m-i)*1LL*a[i].va;
        cout<<sum;
        for(int i=0;i<m;i++)cout<<" "<<a[i].s;
        cout<<"\n";
    }
    return 0;
}

Lucky 7

---

Time Limit: 1 Second       Memory Limit: 65536 KB

---

BaoBao has just found a positive integer sequence  of length  from his left pocket and another positive integer  from his right pocket. As number 7 is BaoBao''s favorite number, he considers a positive integer  lucky if  is divisible by 7. He now wants to select an integer  from the sequence such that  is lucky. Please tell him if it is possible.

Input

There are multiple test cases. The first line of the input is an integer  (about 100), indicating the number of test cases. For each test case:

The first line contains two integers  and  (), indicating the length of the sequence and the positive integer in BaoBao''s right pocket.

The second line contains  positive integers  (), indicating the sequence.

Output

For each test case output one line. If there exists an integer  such that  and  is lucky, output "Yes" (without quotes), otherwise output "No" (without quotes).

Sample Input

4
3 7
4 5 6
3 7
4 7 6
5 2
2 5 2 5 2
4 26
100 1 2 4

Sample Output

No
Yes
Yes
Yes

Hint

For the first sample test case, as 4 + 7 = 11, 5 + 7 = 12 and 6 + 7 = 13 are all not divisible by 7, the answer is "No".

For the second sample test case, BaoBao can select a 7 from the sequence to get 7 + 7 = 14. As 14 is divisible by 7, the answer is "Yes".

For the third sample test case, BaoBao can select a 5 from the sequence to get 5 + 2 = 7. As 7 is divisible by 7, the answer is "Yes".

For the fourth sample test case, BaoBao can select a 100 from the sequence to get 100 + 26 = 126. As 126 is divisible by 7, the answer is "Yes".

加上一个数是不是7的倍数，直接做

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int n,b;
        cin>>n>>b;
        int f=0;
        for(int i=0,x;i<n;i++)
        {
            cin>>x;
            if((x+b)%7==0)f=1;
        }
        cout<<(f?"Yes":"No")<<"\n";
    }
    return 0;
}

 

K

Mahjong Sorting

---

Time Limit: 1 Second       Memory Limit: 65536 KB

---

DreamGrid has just found a set of Mahjong with  suited tiles and a White Dragon tile in his pocket. Each suited tile has a suit (Character, Bamboo or Dot) and a rank (ranging from 1 to ), and there is exactly one tile of each rank and suit combination.

Character tiles whose rank ranges from 1 to 9

Bamboo tiles whose rank ranges from 1 to 9

Dot tiles whose rank ranges from 1 to 9

White Dragon tile

As DreamGrid is bored, he decides to play with these tiles. He first selects one of the  suited tiles as the "lucky tile", then he picks  tiles from the set of  tiles and sorts these  tiles with the following rules:

• The "lucky tile", if contained in the  tiles, must be placed in the leftmost position.

• For two tiles  and  such that neither of them is the "lucky tile", if

  •  is a Character tile and  is a Bamboo tile, or

  •  is a Character tile and  is a Dot tile, or

  •  is a Bamboo tile and  is a Dot tile, or

  •  and  have the same suit and the rank of  is smaller than the rank of ,

  then  must be placed to the left of .

   

 

White Dragon tile is a special tile. If it''s contained in the  tiles, it''s considered as the original (not-lucky) version of the lucky tile during the sorting. For example, consider the following sorted tiles, where "3 Character" is selected as the lucky tile. White Dragon tile, in this case, is considered to be the original not-lucky version of "3 Character" and should be placed between "2 Character" and "4 Character".

As DreamGrid is quite forgetful, he immediately forgets what the lucky tile is after the sorting! Given  sorted tiles, please tell DreamGrid the number of possible lucky tiles.

Input

There are multiple test cases. The first line of the input contains an integer , indicating the number of test cases. For each test case:

The first line contains two integers  and  (, ), indicating the number of sorted tiles and the maximum rank of suited tiles.

For the next  lines, the -th line describes the -th sorted tile counting from left to right. The line begins with a capital letter  (), indicating the suit of the -th tile:

• If , then an integer  () follows, indicating that it''s a Character tile with rank ;

• If , then an integer  () follows, indicating that it''s a Bamboo tile with rank ;

• If , then an integer  () follows, indicating that it''s a Dot tile with rank ;

• If , then it''s a White Drangon tile.

 

It''s guaranteed that there exists at least one possible lucky tile, and the sum of  in all test cases doesn''t exceed .

Output

For each test case output one line containing one integer, indicating the number of possible lucky tiles.

Sample Input

4
3 9
C 2
W
C 4
6 9
C 2
C 7
W
B 3
B 4
D 2
3 100
C 2
W
C 9
3 9
C 1
B 2
D 3

Sample Output

2
4
7
25

Hint

For the first sample, "2 Character" and "3 Character" are possible lucky tiles.

For the second sample, "8 Character", "9 Character", "1 Bamboo" and "2 Bamboo" are possible lucky tiles.

 自己菜做不出来，要把所有的情况都分析到，我太菜了，这个题不想再做了

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int a[N],n,m;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int f=0,ans;
        scanf("%d%d",&n,&m);
        a[n+1]=3*m+1;
        for(int i=1; i<=n; i++)
        {
            getchar();
            char c=getchar();
            if(c!=''W'')
            {
                scanf("%d",&a[i]);
                if(c==''B'')a[i]+=m;
                if(c==''D'')a[i]+=2*m;
            }
            else f=i;
        }
        if(n==1)ans=3*m;
        else
        {
            if(f==0)
                if(a[1]>a[2])ans=1;
                else ans=3*m-n+1;
            else
            {
                if(f==1)ans=a[2]-1;
                else
                {
                    if(a[1]>a[2]&&f!=2)ans=1;
                    else ans=a[f+1]-a[f-1]-(f!=2);
                }
            }
        }
        cout<<ans<<"\n";
    }
    return 0;
}

 

 D

Sequence Swapping

---

Time Limit: 1 Second       Memory Limit: 65536 KB

---

BaoBao has just found a strange sequence {<, >, <, >, , <, >} of length  in his pocket. As you can see, each element <, > in the sequence is an ordered pair, where the first element  in the pair is the left parenthesis ''('' or the right parenthesis '')'', and the second element  in the pair is an integer.

As BaoBao is bored, he decides to play with the sequence. At the beginning, BaoBao''s score is set to 0. Each time BaoBao can select an integer , swap the -th element and the -th element in the sequence, and increase his score by , if and only if ,  ''('' and  '')''.

BaoBao is allowed to perform the swapping any number of times (including zero times). What''s the maximum possible score BaoBao can get?

Input

There are multiple test cases. The first line of the input contains an integer , indicating the number of test cases. For each test case:

The first line contains an integer  (), indicating the length of the sequence.

The second line contains a string  () consisting of ''('' and '')''. The -th character in the string indicates , of which the meaning is described above.

The third line contains  integers  (). Their meanings are described above.

It''s guaranteed that the sum of  of all test cases will not exceed .

Output

For each test case output one line containing one integer, indicating the maximum possible score BaoBao can get.

Sample Input

4
6
)())()
1 3 5 -1 3 2
6
)())()
1 3 5 -100 3 2
3
())
1 -1 -1
3
())
-1 -1 -1

Sample Output

24
21
0
2

Hint

For the first sample test case, the optimal strategy is to select  in order.

For the second sample test case, the optimal strategy is to select  in order.

 这个dp就是去找(，找到它最右可以到达的位置

 然后你可以找一个这个)前一个）,保证他也最优 

#include<bits/stdc++.h>
using namespace std;
const int N=1005;
typedef long long ll;
char s[N];
ll v[N],dp[N][N],sum[N],F[N],nxt[N];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,f=0,tot=1;
        scanf("%d%s",&n,s+1);
        for(int i=1; i<=n; i++)scanf("%lld",&v[i]);
        for(int i=n; i>0; i--)if(s[i]==''('')F[tot++]=i;
        for(int i=1; i<=n; i++) sum[i]=sum[i-1]+(s[i]=='')''?v[i]:0);
        memset(nxt,0,sizeof(nxt));
        for(int i=1; i<=n; i++)
            if(s[i]=='')'')
            {
                if(f) nxt[f]=i;
                f=i;
            }
        for(int i=1; i<=n; i++) dp[i][0]=-(1LL<<60);
        ll ans=0;
        for(int i=1; i<tot; i++)
        {
            for(int j=F[i]+1; j<=n; j++)
                if(s[j]=='')'') dp[i][j]=v[F[i]]*(sum[j]-sum[F[i]])+dp[i-1][j];
            for(int j=n; j>=F[i]; j--)
                if(s[j]=='')'') dp[i][j]=max(dp[i][nxt[j]],dp[i][j]);
            for(int j=F[i]-1; j>=1; j--)
                if(s[j]=='')'') dp[i][j]=max(dp[i-1][j],dp[i][nxt[j]]);
            for(int j=1; j<=n; j++) if(s[j]=='')'')ans=max(ans,dp[i][j]);
        }
        printf("%lld\n",ans);
    }
    return 0;
}

使用滚动数组优化

#include<bits/stdc++.h>
using namespace std;
const int N=1005;
typedef long long ll;
char s[N];
ll v[N],dp[2][N],sum[N],F[N],nxt[N];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,f=0,tot=1,cur=0;
        scanf("%d%s",&n,s+1);
        for(int i=1; i<=n; i++)scanf("%lld",&v[i]);
        for(int i=n; i>0; i--)if(s[i]==''('')F[tot++]=i;
        for(int i=1; i<=n; i++) sum[i]=sum[i-1]+(s[i]=='')''?v[i]:0);
        memset(nxt,0,sizeof(nxt));
        for(int i=1; i<=n; i++)
            if(s[i]=='')'')
            {
                if(f) nxt[f]=i;
                f=i;
            }
        for(int i=1; i<=n; i++) dp[1][i]=dp[0][i]=0;
        dp[0][0]=dp[1][0]=-(1LL<<60);
        ll ans=0;
        for(int i=1; i<tot; i++)
        {
            for(int j=F[i]+1; j<=n; j++)
                if(s[j]=='')'') dp[cur][j]=v[F[i]]*(sum[j]-sum[F[i]])+dp[cur^1][j];
            for(int j=n; j>=F[i]; j--)
                if(s[j]=='')'') dp[cur][j]=max(dp[cur][nxt[j]],dp[cur][j]);
            for(int j=F[i]-1; j>=1; j--)
                if(s[j]=='')'') dp[cur][j]=max(dp[cur^1][j],dp[cur][nxt[j]]);
            for(int j=1; j<=n; j++) if(s[j]=='')'')ans=max(ans,dp[cur][j]);
            cur^=1;
        }
        printf("%lld\n",ans);
    }
    return 0;

 

Magic 12 Months

---

Time Limit: 1 Second       Memory Limit: 65536 KB

---

It''s New Year''s Eve, and it''s also the best time of the year to play the card game Magic 12 Months to pray for good luck of the coming year. BaoBao has just found a deck of standard 52 playing cards (without Jokers) in his pocket and decides to play the game. The rules are as follows:

1. Setup

   1. Remove the four ''K''s from the 52 cards.

   2. Shuffle the remaining 48 cards and divide them face down into 12 piles (4 cards per pile) with equal probability.

2. Gameplay

   1. Let .

   2. Flip the card on the top of the -th pile, check its rank , and discard the card.

   3. If  is a number, let ; If , let ; If , let ; If , let .

   4. If the -th pile is empty, the game ends; Otherwise go back to step 2.2.

When the game ends, having all the 4 cards of rank  flipped and discarded indicates that the -th month in the coming year is a lucky month.

BaoBao is in the middle of the game and has discarded  cards. He wants to know the probability that the -th month of the coming year is a lucky month for all  when the game ends. Given these  cards, please help him calculate the answer.

Input

There are multiple test cases. The first line of input contains an integer  (about 100) -- the number of test cases. For each test case:

The first and only line contains an integer  () -- the number of flipped cards, followed by the rank of the  cards  () separated by a space in the order they are flipped. It''s guaranteed that the input describes a valid and possible situation of the game.

Output

For each test case output one line containing 12 numbers separated by a space, where the -th number indicates the probability that the -th month of the coming year is a lucky month.

You should output a probability in its simplest fraction form  where  and  are coprime. Specifically, if the probability equals 0, you should output 0; If the probability equals 1, you should output 1.

Please, DO NOT output extra spaces at the end of each line, or your answer may be considered incorrect!

Sample Input

3
30 9 Q 10 J Q 10 J 10 J J 8 5 7 6 5 7 6 7 6 6 3 A 2 4 A 2 4 2 4 4
0
7 2 A 3 A 4 A A

Sample Output

1 2/3 2/5 1 1/2 1 2/3 2/5 2/5 2/3 1 1/2
1 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2
1 0 0 0 0 0 0 0 0 0 0 0

 

 

想到了和第一堆有关，但是还没推出来

#include<bits/stdc++.h>
using namespace std;
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
typedef long long ll;
ll C(int n,int m)
{
    if(m>n)return 0;
    ll ans=1;
    for(int i=n-m+1; i<=n; i++)ans*=i;
    for(int i=2; i<=m; i++)ans/=i;
    return ans;
}
int a[15];
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        for(int i=1; i<=12; i++)a[i]=4;
        int n;
        cin>>n;
        string s;
        for(int i=0,x; i<n; i++)
        {
            cin>>s;
            if(s=="10")x=10;
            else if(s=="A")x=1;
            else if(s=="J")x=11;
            else if(s=="Q")x=12;
            else x=s[0]-48;
            a[x]--;
        }
        cout<<1;
        for(int i=2; i<=12; i++)
        {
            //dbg(a[i]);
            if(!a[i])cout<<" 1";
            else if(!a[1])cout<<" 0";
            else
            {
                int m=48-n;
                ll B=C(m,a[1])*C(m-a[1],a[i]),A=0;
                for(int j=a[1]; j<=m; j++)A+=C(j-1,a[1]-1)*C(j-a[1],a[i]);
                ll d=__gcd(A,B);
                if(!A)cout<<" 0";
                else if(A==B)cout<<" 1";
                else cout<<" "<<A/d<<"/"<<B/d;
            }
        }
        cout<<"\n";
    }
    return 0;
}