此处将为大家介绍关于EducationalCodeforcesRound60(RatedforDiv.2)D.MagicGems的详细内容,并且为您解答有关矩阵快速幂的相关问题,此外,我们还将为您介绍
此处将为大家介绍关于Educational Codeforces Round 60 (Rated for Div. 2) D. Magic Gems的详细内容,并且为您解答有关矩阵快速幂的相关问题,此外,我们还将为您介绍关于Coloring Edges(有向图环染色)-- Educational Codeforces Round 72 (Rated for Div. 2)、Educational Codeforces Round 33 (Rated for Div. 2)、Educational Codeforces Round 35 (Rated for Div. 2)、Educational Codeforces Round 35 (Rated for Div. 2)A,B,C,D的有用信息。
本文目录一览:- Educational Codeforces Round 60 (Rated for Div. 2) D. Magic Gems(矩阵快速幂)(矩阵快速幂是什么)
- Coloring Edges(有向图环染色)-- Educational Codeforces Round 72 (Rated for Div. 2)
- Educational Codeforces Round 33 (Rated for Div. 2)
- Educational Codeforces Round 35 (Rated for Div. 2)
- Educational Codeforces Round 35 (Rated for Div. 2)A,B,C,D
Educational Codeforces Round 60 (Rated for Div. 2) D. Magic Gems(矩阵快速幂)(矩阵快速幂是什么)
题目传送门
题意:
一个魔法水晶可以分裂成m个水晶,求放满n个水晶的方案数(mol1e9+7)
思路:
线性dp,dp[i]=dp[i]+dp[i-m];
由于n到1e18,所以要用到矩阵快速幂优化
注意初始化
代码:
#include<bits/stdc++.h> using namespace std; #define mod 1000000007 typedef long long ll; #define MAX 105 const int N=105;//矩阵的大小 int T; ll n,m; ll add(ll a,ll b) { a%=mod; b%=mod; return (a+b)%mod; } struct hh { ll ma[N][N]; }a,res; hh multi(hh a,hh b) { hh tmp; memset(tmp.ma,0,sizeof(tmp.ma)); for(int i=0;i<N;i++) for(int j=0;j<N;j++) for(int k=0;k<N;k++) { tmp.ma[i][j]=add(tmp.ma[i][j],a.ma[i][k]*b.ma[k][j]); } return tmp; } void fast_pow(hh a,long long k) { memset(res.ma,sizeof(res.ma)); for(int i=0;i<N;i++)res.ma[i][i]=1; while(k>0) { if(k&1) res=multi(res,a); a=multi(a,a); k>>=1; } } int main() { while(~scanf("%lld%d",&n,&m)) { for(int i=1;i<=m;i++) a.ma[i][i-1]=1; a.ma[1][1]=a.ma[1][m]=1; fast_pow(a,n); printf("%lld\n",res.ma[1][1]); } return 0; }
Coloring Edges(有向图环染色)-- Educational Codeforces Round 72 (Rated for Div. 2)
题意:https://codeforc.es/contest/1217/problem/D
给你一个有向图,要求一个循环里不能有相同颜色的边,问你最小要几种颜色染色,怎么染色?
思路:
如果没有环,那全是1;如果有环,那小到大的边为1,大到小的边为2。
1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0); 2 #include <cstdio>//sprintf islower isupper 3 #include <cstdlib>//malloc exit strcat itoa system("cls") 4 #include <iostream>//pair 5 #include <fstream>//freopen("C:\\Users\\13606\\Desktop\\草稿.txt","r",stdin); 6 #include <bitset> 7 //#include <G> 8 //#include<unordered_map> 9 #include <vector> 10 #include <stack> 11 #include <set> 12 #include <string.h>//strstr substr 13 #include <string> 14 #include <time.h>//srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9; 15 #include <cmath> 16 #include <deque> 17 #include <queue>//priority_queue<int,vector<int>,greater<int> > q;//less 18 #include <vector>//emplace_back 19 //#include <math.h> 20 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor 21 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first,nth,last,compare) 22 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation 23 #define fo(a,b,c) for(register int a=b;a<=c;++a) 24 #define fr(a,c) for(register int a=b;a>=c;--a) 25 #define mem(a,b) memset(a,sizeof(a)) 26 #define pr printf 27 #define sc scanf 28 #define ls rt<<1 29 #define rs rt<<1|1 30 typedef long long ll; 31 void swapp(int &a,int &b); 32 double fabss(double a); 33 int maxx(int a,int b); 34 int minn(int a,int b); 35 int Del_bit_1(int n); 36 int lowbit(int n); 37 int abss(int a); 38 //const long long INF=(1LL<<60); 39 const double E=2.718281828; 40 const double PI=acos(-1.0); 41 const int inf=(1<<30); 42 const double ESP=1e-9; 43 const int mod=(int)1e9+7; 44 const int N=(int)1e6+10; 45 46 int in[N]; 47 vector<vector<int> > G(N); 48 49 bool top_sort(int n) 50 { 51 int cont=0; 52 queue<int> q; 53 for(int i=1;i<=n;i++) 54 if(in[i]==0) 55 q.push(i); 56 while(!q.empty()) 57 { 58 int x=q.front(); 59 q.pop(); 60 cont++; 61 for(int i=0;i<G[x].size();i++) 62 { 63 in[G[x][i]]--; 64 if(in[G[x][i]]==0) 65 q.push(G[x][i]); 66 } 67 } 68 return (cont==n); 69 } 70 struct node 71 { 72 int u,v; 73 }edge[N]; 74 75 int main() 76 { 77 int n,m; 78 sc("%d%d",&n,&m); 79 for(int i=1;i<=m;++i) 80 { 81 int u,v; 82 sc("%d%d",&u,&v); 83 edge[i]={u,v}; 84 in[v]++; 85 G[u].push_back(v); 86 } 87 if(top_sort(n)) 88 { 89 pr("1\n"); 90 for(int i=1;i<=m;++i) 91 pr("1 "); 92 } 93 else 94 { 95 pr("2\n"); 96 for(int i=1;i<=m;++i) 97 pr("%d ",edge[i].u>edge[i].v?1:2); 98 } 99 return 0; 100 } 101 102 /**************************************************************************************/ 103 104 int maxx(int a,int b) 105 { 106 return a>b?a:b; 107 } 108 109 void swapp(int &a,int &b) 110 { 111 a^=b^=a^=b; 112 } 113 114 int lowbit(int n) 115 { 116 return n&(-n); 117 } 118 119 int Del_bit_1(int n) 120 { 121 return n&(n-1); 122 } 123 124 int abss(int a) 125 { 126 return a>0?a:-a; 127 } 128 129 double fabss(double a) 130 { 131 return a>0?a:-a; 132 } 133 134 int minn(int a,int b) 135 { 136 return a<b?a:b; 137 }
Educational Codeforces Round 33 (Rated for Div. 2)
总结
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Educational Codeforces Round 35 (Rated for Div. 2)
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Educational Codeforces Round 35 (Rated for Div. 2)A,B,C,D
You are given an array of n integer numbers a0, a1, ..., an - 1. Find the distance between two closest (nearest) minimums in it. It is guaranteed that in the array a minimum occurs at least two times.
The first line contains positive integer n (2 ≤ n ≤ 105) — size of the given array. The second line contains n integers a0, a1, ..., an - 1(1 ≤ ai ≤ 109) — elements of the array. It is guaranteed that in the array a minimum occurs at least two times.
Print the only number — distance between two nearest minimums in the array.
2
3 3
1
3
5 6 5
2
9
2 1 3 5 4 1 2 3 1
3
思路:水题
实现代码:
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n,minn = 1100000002,a[200000];
cin>>n;
for(int i = 0;i < n;i ++){
cin>>a[i];
minn = min(minn , a[i]);
}
int flag = 0;
int maxx = 1100000002,ans=1;
for(int i = 0;i < n;i ++){
if(a[i] == minn){
if(flag)
maxx = min(ans,maxx);
flag = 1;
ans = 1;
}
else ans++;
}
cout<<maxx<<endl;
}
It''s New Year''s Eve soon, so Ivan decided it''s high time he started setting the table. Ivan has bought two cakes and cut them into pieces: the first cake has been cut into a pieces, and the second one — into b pieces.
Ivan knows that there will be n people at the celebration (including himself), so Ivan has set n plates for the cakes. Now he is thinking about how to distribute the cakes between the plates. Ivan wants to do it in such a way that all following conditions are met:
- Each piece of each cake is put on some plate;
- Each plate contains at least one piece of cake;
- No plate contains pieces of both cakes.
To make his guests happy, Ivan wants to distribute the cakes in such a way that the minimum number of pieces on the plate is maximized. Formally, Ivan wants to know the maximum possible number x such that he can distribute the cakes according to the aforementioned conditions, and each plate will contain at least x pieces of cake.
Help Ivan to calculate this number x!
The first line contains three integers n, a and b (1 ≤ a, b ≤ 100, 2 ≤ n ≤ a + b) — the number of plates, the number of pieces of the first cake, and the number of pieces of the second cake, respectively.
Print the maximum possible number x such that Ivan can distribute the cake in such a way that each plate will contain at least x pieces of cake.
5 2 3
1
4 7 10
3
In the first example there is only one way to distribute cakes to plates, all of them will have 1 cake on it.
In the second example you can have two plates with 3 and 4 pieces of the first cake and two plates both with 5 pieces of the second cake. Minimal number of pieces is 3.
思路:
水题
实现代码:
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n,a,b;
cin>>n>>a>>b;
int ans = 0;
for(int i = 1;i <= n-1;i++){
ans = max(ans,min(a/i,b/(n-i)));
}
cout<<ans<<endl;
}
Mishka is decorating the Christmas tree. He has got three garlands, and all of them will be put on the tree. After that Mishka will switch these garlands on.
When a garland is switched on, it periodically changes its state — sometimes it is lit, sometimes not. Formally, if i-th garland is switched on during x-th second, then it is lit only during seconds x, x + ki, x + 2ki, x + 3ki and so on.
Mishka wants to switch on the garlands in such a way that during each second after switching the garlands on there would be at least one lit garland. Formally, Mishka wants to choose three integers x1, x2 and x3 (not necessarily distinct) so that he will switch on the first garland during x1-th second, the second one — during x2-th second, and the third one — during x3-th second, respectively, and during each second starting from max(x1, x2, x3) at least one garland will be lit.
Help Mishka by telling him if it is possible to do this!
The first line contains three integers k1, k2 and k3 (1 ≤ ki ≤ 1500) — time intervals of the garlands.
If Mishka can choose moments of time to switch on the garlands in such a way that each second after switching the garlands on at least one garland will be lit, print YES.
Otherwise, print NO.
2 2 3
YES
4 2 3
NO
In the first example Mishka can choose x1 = 1, x2 = 2, x3 = 1. The first garland will be lit during seconds 1, 3, 5, 7, ..., the second — 2, 4, 6, 8, ..., which already cover all the seconds after the 2-nd one. It doesn''t even matter what x3 is chosen. Our choice will lead third to be lit during seconds 1, 4, 7, 10, ..., though.
In the second example there is no way to choose such moments of time, there always be some seconds when no garland is lit.
思路:
1 = 1/2 + 1/2 ,1 = 1/2+1/4+1/4, 差不多就这意思
实现代码:
#include<bits/stdc++.h>
using namespace std;
int isPrime(int n) {
int i;
for (i = 2; i * i <= n; ++i) {
if (n % i == 0) return 0;
}
return 1;
}
int main()
{
int a[5];
double ans = 0;
for(int i = 0;i < 3;i ++)
cin>>a[i];
sort(a,a+3);
for(int i = 2;i <= 1500;i++){
if(isPrime(i)){
ans = 0;
for(int j = 0;j < 3;j ++){
if(a[j]==1){
cout<<"YES"<<endl; return 0;
}
if(a[j]%i==0){
ans += 1.0/a[j];
}
if(ans == 1){
cout<<"YES"<<endl;return 0;}
}
}
}
cout<<"NO"<<endl;
}
A permutation of size n is an array of size n such that each integer from 1 to n occurs exactly once in this array. An inversion in a permutation p is a pair of indices (i, j) such that i > j and ai < aj. For example, a permutation [4, 1, 3, 2] contains 4 inversions: (2, 1), (3, 1), (4, 1), (4, 3).
You are given a permutation a of size n and m queries to it. Each query is represented by two indices l and r denoting that you have to reverse the segment [l, r] of the permutation. For example, if a = [1, 2, 3, 4] and a query l = 2, r = 4 is applied, then the resulting permutation is [1, 4, 3, 2].
After each query you have to determine whether the number of inversions is odd or even.
The first line contains one integer n (1 ≤ n ≤ 1500) — the size of the permutation.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ n) — the elements of the permutation. These integers are pairwise distinct.
The third line contains one integer m (1 ≤ m ≤ 2·105) — the number of queries to process.
Then m lines follow, i-th line containing two integers li, ri (1 ≤ li ≤ ri ≤ n) denoting that i-th query is to reverse a segment [li, ri] of the permutation. All queries are performed one after another.
Print m lines. i-th of them must be equal to odd if the number of inversions in the permutation after i-th query is odd, and even otherwise.
3
1 2 3
2
1 2
2 3
odd
even
4
1 2 4 3
4
1 1
1 4
1 4
2 3
odd
odd
odd
even
The first example:
- after the first query a = [2, 1, 3], inversion: (2, 1);
- after the second query a = [2, 3, 1], inversions: (3, 1), (3, 2).
The second example:
- a = [1, 2, 4, 3], inversion: (4, 3);
- a = [3, 4, 2, 1], inversions: (3, 1), (4, 1), (3, 2), (4, 2), (4, 3);
- a = [1, 2, 4, 3], inversion: (4, 3);
- a = [1, 4, 2, 3], inversions: (3, 2), (4, 2).
思路:和 l,r 的值没关系,只和他们之间的距离有关系
实现代码:
#include<bits/stdc++.h>
using namespace std;
const int M = 2e5+100;
int a[M],t[M];
void merge_sort(int x, int y, int & ans){
if(y-x>1){
int m = x+(y-x)/2;
int p=x, q=m, i=x;
merge_sort(x, m, ans);
merge_sort(m ,y, ans);
while(p<m || q<y){
if(q>=y || p<m && a[p]<=a[q]){
t[i++] = a[p++];
}
else{
t[i++] = a[q++];
ans += m-p;
}
}
for(int j=x; j<y; j++){
a[j] = t[j];
}
}
}
int main(){
int n,q,l,r;
cin>>n;
for(int i = 0;i < n; i++){
cin>>a[i];
}
int ans = 0;
merge_sort(0,n,ans);
if(ans % 2 == 0)
ans = 0;
else
ans = 1;
cin>>q;
while(q--){
cin>>l>>r;
int len = r - l + 1;
len/=2;
if(len % 2 == 0){
if(ans == 0) cout<<"even"<<endl;
else cout<<"odd"<<endl;
}
else{
if(ans == 0){ ans = 1; cout<<"odd"<<endl;}
else{ ans = 0; cout<<"even"<<endl;}
}
}
return 0;
}
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