#include <bits/stdc++.h> using namespace std; #define ll long long #define TLE ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); ll m,n,l,r; int main() { TLE; while(cin>>n>>m>>l>>r) { for(int i=0;i<min(n,l);i++) { if(m-i*r<0) break; if( !((m-i*r)%l) ) { if(i+((m-i*r)/l) <=n ) { cout<<(m-i*r)/l<<" "<<i<<" "<<n-i-(m-i*r)/l<<endl; return 0; } } } cout<<-1<<endl; } return 0; }

Today''s morning was exceptionally snowy. Meshanya decided to go outside and noticed a huge snowball rolling down the mountain! Luckily, there are two stones on that mountain.

Initially, snowball is at height hh and it has weight ww. Each second the following sequence of events happens: snowball''s weights increases by ii, where ii — is the current height of snowball, then snowball hits the stone (if it''s present at the current height), then snowball moves one meter down. If the snowball reaches height zero, it stops.

There are exactly two stones on the mountain. First stone has weight u1u1 and is located at height d1d1, the second one — u2u2 and d2d2respectively. When the snowball hits either of two stones, it loses weight equal to the weight of that stone. If after this snowball has negative weight, then its weight becomes zero, but the snowball continues moving as before.

Find the weight of the snowball when it stops moving, that is, it reaches height 0.

First line contains two integers ww and hh — initial weight and height of the snowball (0≤w≤1000≤w≤100; 1≤h≤1001≤h≤100).

Second line contains two integers u1u1 and d1d1 — weight and height of the first stone (0≤u1≤1000≤u1≤100; 1≤d1≤h1≤d1≤h).

Third line contains two integers u2u2 and d2d2 — weight and heigth of the second stone (0≤u2≤1000≤u2≤100; 1≤d2≤h1≤d2≤h; d1≠d2d1≠d2). Notice that stones always have different heights.

Output a single integer — final weight of the snowball after it reaches height 0.

In the first example, initially a snowball of weight 4 is located at a height of 3, there are two stones of weight 1, at a height of 1 and 2, respectively. The following events occur sequentially:

• The weight of the snowball increases by 3 (current height), becomes equal to 7.
• The snowball moves one meter down, the current height becomes equal to 2.
• The weight of the snowball increases by 2 (current height), becomes equal to 9.
• The snowball hits the stone, its weight decreases by 1 (the weight of the stone), becomes equal to 8.
• The snowball moves one meter down, the current height becomes equal to 1.
• The weight of the snowball increases by 1 (current height), becomes equal to 9.
• The snowball hits the stone, its weight decreases by 1 (the weight of the stone), becomes equal to 8.
• The snowball moves one meter down, the current height becomes equal to 0.

Thus, at the end the weight of the snowball is equal to 8.

You are given an undirected graph consisting of n vertices and  edges. Instead of giving you the edges that exist in the graph, we give you m unordered pairs (x, y) such that there is no edge between x and y, and if some pair of vertices is not listed in the input, then there is an edge between these vertices.

You have to find the number of connected components in the graph and the size of each component. A connected component is a set of vertices X such that for every two vertices from this set there exists at least one path in the graph connecting these vertices, but adding any other vertex to X violates this rule.

The first line contains two integers n and m (1 ≤ n ≤ 200000, ).

Then m lines follow, each containing a pair of integers x and y (1 ≤ x, y ≤ n, x ≠ y) denoting that there is no edge between x and y. Each pair is listed at most once; (x, y) and (y, x) are considered the same (so they are never listed in the same test). If some pair of vertices is not listed in the input, then there existsan edge between those vertices.

Firstly print k — the number of connected components in this graph.

Then print k integers — the sizes of components. You should output these integers in non-descending order.

 1 /*
 2 Welcome Hacking
 3 Wish You High Rating
 4 */
 5 #include<iostream>
 6 #include<cstdio>
 7 #include<cstring>
 8 #include<ctime>
 9 #include<cstdlib>
10 #include<algorithm>
11 #include<cmath>
12 #include<string>
13 #include<map>
14 #include<set>
15 using namespace std;
16 int read(){
17     int xx=0,ff=1;char ch=getchar();
18     while(ch>''9''||ch<''0''){if(ch==''-'')ff=-1;ch=getchar();}
19     while(ch>=''0''&&ch<=''9''){xx=(xx<<3)+(xx<<1)+ch-''0'';ch=getchar();}
20     return xx*ff;
21 }
22 const int maxn=200010;
23 int N,M,t1,t2,temp[maxn],ans;
24 set<int>s;
25 map<int,bool>e[maxn];
26 void dfs(int x){
27     int prev=0;
28     while(1){
29         int t=*s.lower_bound(prev);
30         if(t==(1<<30))
31             break;
32         prev=t+1;
33         if(!e[x][t]){
34             temp[ans]++;
35             s.erase(t);
36             dfs(t);
37         }
38     }
39 }
40 int main(){
41     //freopen("in","r",stdin);
42     N=read(),M=read();
43     for(int i=1;i<=N;i++)
44         s.insert(i);
45     s.insert(1<<30);
46     for(int i=1;i<=M;i++){
47         t1=read(),t2=read();
48         e[t1][t2]=e[t2][t1]=1;
49     }
50     int prev=0;
51     while(1){
52         int t=*s.lower_bound(prev);
53         if(t==(1<<30))
54             break;
55         prev=t+1;
56         s.erase(t);
57         temp[++ans]=1;
58         dfs(t);
59     }
60     sort(temp+1,temp+1+ans);
61     printf("%d\n",ans);
62     for(int i=1;i<=ans;i++)
63         printf("%d ",temp[i]);
64     puts("");
65     return 0;
66 }

 1 /*
 2 Welcome Hacking
 3 Wish You High Rating
 4 */
 5 #include<iostream>
 6 #include<cstdio>
 7 #include<cstring>
 8 #include<ctime>
 9 #include<cstdlib>
10 #include<algorithm>
11 #include<cmath>
12 #include<string>
13 using namespace std;
14 inline int read(){
15     int xx=0,ff=1;char ch=getchar();
16     while(ch>''9''||ch<''0''){if(ch==''-'')ff=-1;ch=getchar();}
17     while(ch>=''0''&&ch<=''9''){xx=(xx<<3)+(xx<<1)+ch-''0'';ch=getchar();}
18     return xx*ff;
19 }
20 inline int mymin(int xx,int yy)
21 {if(xx<yy)return xx;return yy;}
22 const int maxn=300010,limit=1000000;
23 int N,M,D[limit+10],tot=6,sum[maxn],a[maxn];
24 struct Segment_Tree{
25     long long sum;
26 }T[maxn*4];
27 int x,y,opt;
28 void build(int L,int R,int root){
29     if(L==R){
30         T[root].sum=a[L];
31         return;
32     }
33     int mid=(L+R)>>1;
34     build(L,mid,root*2);
35     build(mid+1,R,root*2+1);
36     T[root].sum=T[root*2].sum+T[root*2+1].sum;
37 }
38 void upd(int L,int R,int root){
39     if(T[root].sum==(R-L+1)*2)
40         return;
41     if(x>R||y<L)
42         return;
43     if(L==R){
44         T[root].sum=D[T[root].sum];
45         return;
46     }
47     int mid=(L+R)>>1;
48     upd(L,mid,root*2);
49     upd(mid+1,R,root*2+1);
50     T[root].sum=T[root*2].sum+T[root*2+1].sum;
51 }
52 long long query(int L,int R,int root){
53     if(x>R||y<L)
54         return 0;
55     if(x<=L&&y>=R)
56         return T[root].sum;
57     int mid=(L+R)>>1;
58     return query(L,mid,root*2)+query(mid+1,R,root*2+1);
59 }
60 int main(){
61     //freopen("in","r",stdin);
62     for(int i=1;i<=limit;i++)
63         for(int j=i;j<=limit;j+=i)
64             D[j]++;
65     N=read(),M=read();
66     for(int i=1;i<=N;i++){
67         a[i]=read();
68         sum[i]=sum[i-1];
69         if(a[i]==1)
70             a[i]=2,sum[i]++;
71     }
72     build(1,N,1);
73     while(M--){
74         opt=read();
75         if(opt==1){
76             x=read(),y=read();
77             upd(1,N,1);
78         }
79         else{
80             x=read(),y=read();
81             printf("%I64d\n",query(1,N,1)-(sum[y]-sum[x-1]));
82         }
83     }
84     return 0;
85 }

 1 /*
 2 Welcome Hacking
 3 Wish You High Rating
 4 */
 5 #include<iostream>
 6 #include<cstdio>
 7 #include<cstring>
 8 #include<ctime>
 9 #include<cstdlib>
10 #include<algorithm>
11 #include<cmath>
12 #include<string>
13 #include<vector>
14 using namespace std;
15 int read(){
16     int xx=0,ff=1;char ch=getchar();
17     while(ch>''9''||ch<''0''){if(ch==''-'')ff=-1;ch=getchar();}
18     while(ch>=''0''&&ch<=''9''){xx=(xx<<3)+(xx<<1)+ch-''0'';ch=getchar();}
19     return xx*ff;
20 }
21 const int limit=1000000;
22 vector<int>V[limit+5];
23 void get_fac(){
24     for(int i=2;i<=limit;i++)
25         if(!V[i].size())
26             for(int j=i;j<=limit;j+=i)
27                 V[j].push_back(i);
28 }
29 int x,p,k,delta;
30 int L,R,mid;
31 int calc(int num,int rb){
32     int siz=V[num].size();
33     int re=0;
34     for(int i=1;i<(1<<siz);i++){
35         int s=1,tim=0;
36         for(int j=0;j<siz;j++)
37             if((i>>j)&1)
38                 s*=V[num][j],tim++;
39         if(tim&1)
40             re+=rb/s;
41         else
42             re-=rb/s;
43     }
44     return rb-re;
45 }
46 int main(){
47     //freopen("in","r",stdin);
48     get_fac();
49     for(int T=read();T;T--){
50         x=read(),p=read(),k=read();
51         delta=calc(p,x);
52         L=x,R=int(1e8);
53         while(L+1<R){
54             mid=(L+R)/2;
55             if(calc(p,mid)-delta>=k)
56                 R=mid;
57             else
58                 L=mid;
59         }
60         printf("%d\n",R);
61     }
62     return 0;
63 }