There are three friend living on the straight line Ox in Lineland. The first friend lives at the point x₁, the second friend lives at the point x₂, and the third friend lives at the point x₃. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?

It's guaranteed that the optimal answer is always integer.

Print one integer — the minimum total distance the friends need to travel in order to meet together.

In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.

// <A.cpp> - Mon Oct  3 19:17:39 2016
// This file is made by YJinpeng，created by XuYike's black technology automatically.
// Copyright (C) 2016 ChangJun High School, Inc. // I don't know what this program is.  #include <iostream> #include <vector> #include <algorithm> #include <cstring> #include <cstdio> #include <cstdlib> #include <cmath> #pragma GCC push_options #pragma GCC optimize ("O2") #define MOD 1000000007 #define INF 1e9 #define IN inline #define RG register using namespace std; typedef long long LL; typedef long double LB; const int MAXN=100010; const int MAXM=100010; inline int max(int &x,int &y) {return x>y?x:y;} inline int min(int &x,int &y) {return x<y?x:y;} inline LL gi() { register LL w=0,q=0;register char ch=getchar(); while((ch<'0'||ch>'9')&&ch!='-')ch=getchar(); if(ch=='-')q=1,ch=getchar(); while(ch>='0'&&ch<='9')w=w*10+ch-'0',ch=getchar(); return q?-w:w; } int main() { freopen("A.in","r",stdin); freopen("A.out","w",stdout); int a=gi(),b=gi(),c=gi(); printf("%d",max(max(abs(a-b),abs(b-c)),abs(a-c))); return 0; }

Print two space-separated integers:

• the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
• the number of words inside the parentheses (print 0, if there is no word inside the parentheses).

Note

In the first sample, the words "Hello", "Vasya" and "bye" are outside any of the parentheses, and the words "and", "Petya", "and" and "OK" are inside. Note, that the word "and" is given twice and you should count it twice in the answer.

// <B.cpp> - Mon Oct  3 19:17:39 2016
// This file is made by YJinpeng，created by XuYike's black technology automatically.
// Copyright (C) 2016 ChangJun High School, Inc. // I don't know what this program is.  #include <iostream> #include <vector> #include <algorithm> #include <cstring> #include <cstdio> #include <cstdlib> #include <cmath> using namespace std; const int MAXN=100010; const int MAXM=100010; inline int gi() { register int w=0,q=0;register char ch=getchar(); while((ch<'0'||ch>'9')&&ch!='-')ch=getchar(); if(ch=='-')q=1,ch=getchar(); while(ch>='0'&&ch<='9')w=w*10+ch-'0',ch=getchar(); return q?-w:w; } char s[MAXN]; int main() { freopen("B.in","r",stdin); freopen("B.out","w",stdout); int n=gi(),ans=0,x=0;scanf("%s",s); for(int i=0;i<n;i++){ int j=i; while(s[j]!='('&&s[j]!='_'&&j<n)j++; if(j!=i)ans=max(j-i,ans); if(s[j]=='('){ for(j++;j<n;j++){ if(s[j]==')')break; int o=j; while(s[o]!='_'&&s[o]!=')'&&o<n)o++; if(o!=j)x++,j=o; if(s[o]==')')break; } } i=j; } printf("%d %d",ans,x); return 0; }

Polycarp is a music editor at the radio station. He received a playlist for tomorrow, that can be represented as a sequence a₁, a₂, ..., a_n, where a_i is a band, which performs the i-th song. Polycarp likes bands with the numbers from 1 to m, but he doesn't really like others.

We define as b_j the number of songs the group j is going to perform tomorrow. Polycarp wants to change the playlist in such a way that the minimum among the numbersb₁, b₂, ..., b_m will be as large as possible.

Find this maximum possible value of the minimum among the b_j (1 ≤ j ≤ m), and the minimum number of changes in the playlist Polycarp needs to make to achieve it. One change in the playlist is a replacement of the performer of the i-th song with any other group.

In the first line print two integers: the maximum possible value of the minimum among the b_j(1 ≤ j ≤ m), where b_j is the number of songs in the changed playlist performed by the j-th band, and the minimum number of changes in the playlist Polycarp needs to make.

In the second line print the changed playlist.

If there are multiple answers, print any of them.

// <C.cpp> - Mon Oct  3 19:17:39 2016
// This file is made by YJinpeng，created by XuYike's black technology automatically.
// Copyright (C) 2016 ChangJun High School, Inc. // I don't know what this program is.  #include <iostream> #include <vector> #include <algorithm> #include <cstring> #include <cstdio> #include <cstdlib> #include <cmath> #include <queue> using namespace std; typedef long long LL; const int MAXN=2010; inline LL gi() { register LL w=0,q=0;register char ch=getchar(); while((ch<'0'||ch>'9')&&ch!='-')ch=getchar(); if(ch=='-')q=1,ch=getchar(); while(ch>='0'&&ch<='9')w=w*10+ch-'0',ch=getchar(); return q?-w:w; } int a[MAXN],f[MAXN],ans; int main() { freopen("C.in","r",stdin); freopen("C.out","w",stdout); int n=gi(),m=gi(),x=n/m; for(int i=1;i<=n;i++){a[i]=gi();if(a[i]<=m)f[a[i]]++;} printf("%d ",x); for(int i=1,j;i<=n;i++) if(a[i]>m) for(j=1;j<=m;j++) if(f[j]<x){ ans++;a[i]=j;f[j]++;break; } for(int i=1;i<=m;i++) if(f[i]<x) for(int j=1;j<=n;j++){ if(f[i]==x)break; if(f[a[j]]>x){ f[a[j]]--;f[i]++;ans++;a[j]=i; } } printf("%d\n",ans); for(int i=1;i<=n;i++)printf("%d ",a[i]); return 0; }

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long #define inf 2147483640 #define Pi acos(-1.0) #define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout); using namespace std; const int maxn=2010; int a[maxn],b[maxn],n,m; int main() { scanf("%d%d",&n,&m); for (int i=1;i<=n;i++) scanf("%d",&a[i]); int t=0; for (int i=1;i<=n;i++) { if (a[i]>m) t++; else b[a[i]]++; } int ans=n/m,cnt=0; for (int i=1;i<=m;i++) if (b[i]<ans) { if (t) { for (int j=1;j<=n;j++) { if (a[j]>m) a[j]=i,t--,b[i]++,cnt++; if (b[i]==ans) break; } if (b[i]==ans) continue; } for (int j=1;j<=n;j++) { if (b[a[j]]>ans) {b[a[j]]--;a[j]=i;b[i]++;cnt++;} if (b[i]==ans) break; } } printf("%d %d\n",ans,cnt); for (int i=1;i<=n;i++) printf("%d ",a[i]); return 0; }

The map of Berland is a rectangle of the size n × m, which consists of cells of size 1 × 1. Each cell is either land or water. The map is surrounded by the ocean.

Lakes are the maximal regions of water cells, connected by sides, which are not connected with the ocean. Formally, lake is a set of water cells, such that it's possible to get from any cell of the set to any other without leaving the set and moving only to cells adjacent by the side, none of them is located on the border of the rectangle, and it's impossible to add one more water cell to the set such that it will be connected with any other cell.

You task is to fill up with the earth the minimum number of water cells so that there will be exactly k lakes in Berland. Note that the initial number of lakes on the map is not less than k.

In the first line print the minimum number of cells which should be transformed from water to land.

In the next n lines print m symbols — the map after the changes. The format must strictly follow the format of the map in the input data (there is no need to print the size of the map). If there are several answers, print any of them.

It is guaranteed that the answer exists on the given data.

// <D.cpp> - Mon Oct  3 19:17:39 2016
// This file is made by YJinpeng，created by XuYike's black technology automatically.
// Copyright (C) 2016 ChangJun High School, Inc.
// I don't know what this program is.

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <queue>
#include <cstdlib>
#include <cmath>
using namespace std;
typedef long long LL;
const int MAXN=51;
inline LL gi() {register LL w=0,q=0;register char ch=getchar();while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();if(ch=='-')q=1,ch=getchar();while(ch>='0'&&ch<='9')w=w*10+ch-'0',ch=getchar();return q?-w:w;
}
bool f[MAXN][MAXN],u[MAXN][MAXN];
int _x[5]={0,0,-1,1},_y[5]={-1,1,0,0};
struct node{int x, y;};
struct bb{int s;vector<node>a;bool operator <(bb b)const{return s<b.s;}
}p[MAXN*MAXN];
queue<node>q;int n,m;
void work(int i,int j){q.push((node){i,j});u[i][j]=true;while(!q.empty()){int x=q.front().x,y=q.front().y;q.pop();for(int o=0,a,b;a=x+_x[o],b=y+_y[o],o<4;o++)if(!f[a][b]&&!u[a][b]&&a>=1&&a<=n&&b>=1&&b<=m)u[a][b]=1,q.push((node){a,b});}
}
int main()
{freopen("D.in","r",stdin);freopen("D.out","w",stdout);n=gi(),m=gi();int k=gi(),tot=0;for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){char ch=getchar();while(ch!='.'&&ch!='*')ch=getchar();f[i][j]=bool(ch=='*');}}for(int i=1;i<=m;i++){if(!f[1][i]&&!u[1][i])work(1,i);if(!f[n][i]&&!u[n][i])work(n,i);}for(int i=1;i<=n;i++){if(!f[i][1]&&!u[i][1])work(i,1);if(!f[i][m]&&!u[i][m])work(i,m);}for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)if(!f[i][j]&&!u[i][j]){q.push((node){i,j});p[++tot].s=1;u[i][j]=true;while(!q.empty()){int x=q.front().x,y=q.front().y;p[tot].a.push_back((node){x,y});q.pop();for(int o=0,a,b;a=x+_x[o],b=y+_y[o],o<4;o++)if(!f[a][b]&&!u[a][b]&&a>=1&&a<=n&&b>=1&&b<=m){u[a][b]=1;q.push((node){a,b});p[tot].s++;}}}sort(p+1,p+1+tot);int ans=0;k=tot-k;for(int i=1;i<=k;i++){for(int to=p[i].a.size(),j=0,x,y;x=p[i].a[j].x,y=p[i].a[j].y,j<to;j++)f[x][y]=1;ans+=p[i].s;}printf("%d\n",ans);for(int i=1;i<=n;i++){for(int j=1;j<=m;j++)putchar(f[i][j]?'*':'.');cout<<endl;}return 0;
}

There are n cities and m two-way roads in Berland, each road connects two cities. It is known that there is no more than one road connecting each pair of cities, and there is no road which connects the city with itself. It is possible that there is no way to get from one city to some other city using only these roads.

The road minister decided to make a reform in Berland and to orient all roads in the country, i.e. to make each road one-way. The minister wants to maximize the number of cities, for which the number of roads that begins in the city equals to the number of roads that ends in it.

For each testset print the maximum number of such cities that the number of roads that begins in the city, is equal to the number of roads that ends in it.

In the next m lines print oriented roads. First print the number of the city where the road begins and then the number of the city where the road ends. If there are several answers, print any of them. It is allowed to print roads in each test in arbitrary order. Each road should be printed exactly once.

// <E.cpp> - Mon Oct  3 19:17:39 2016
// This file is made by YJinpeng，created by XuYike's black technology automatically.
// Copyright (C) 2016 ChangJun High School, Inc. // I don't know what this program is.  #include<bits/stdc++.h> #pragma GCC push_options #pragma GCC optimize ("O2") #define IN inline #define RG register using namespace std; const int MAXN=210; inline int gi() { register int w=0,q=0;register char ch=getchar(); while((ch<'0'||ch>'9')&&ch!='-')ch=getchar(); if(ch=='-')q=1,ch=getchar(); while(ch>='0'&&ch<='9')w=w*10+ch-'0',ch=getchar(); return q?-w:w; } set<int>s[MAXN]; vector<pair<int,int> >ans; IN void dfs(RG int x){ while(s[x].size()){ RG int p=*s[x].begin(); s[x].erase(p);s[p].erase(x); ans.push_back(make_pair(x,p));dfs(p); } } int main() { freopen("E.in","r",stdin); freopen("E.out","w",stdout); int T=gi(); while(T--){ ans.clear(); int n=gi(),m=gi(); while(m--){ int x=gi(),y=gi(); s[x].insert(y),s[y].insert(x); } for(int i=1;i<=n;i++) if(s[i].size()&1)s[n+1].insert(i),s[i].insert(n+1); printf("%d\n",n-s[n+1].size()); for(int i=1;i<=n;i++)dfs(i); for(int i=0,to=ans.size();i<to;i++) if(ans[i].first!=n+1&&ans[i].second!=n+1) printf("%d %d\n",ans[i].first,ans[i].second); } return 0; }

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define File(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)
#define maxn 201 #define maxm 40010 using namespace std; typedef long long llg; int T,n,m,sa[maxm],sb[maxm],ls,du[maxn],ans; bool g[maxn][maxn]; int getint(){ int w=0;bool q=0; char c=getchar(); while((c>'9'||c<'0')&&c!='-') c=getchar(); if(c=='-') c=getchar(),q=1; while(c>='0'&&c<='9') w=w*10+c-'0',c=getchar(); return q?-w:w; } void dfs(int u){ while(du[u]){ for(int i=1;i<=n;i++) if(g[u][i]){ ls++; sa[ls]=u; sb[ls]=i; du[u]--,du[i]--; g[u][i]=g[i][u]=0; u=i; break; } } } int main(){ //File("a"); T=getint(); while(T--){ ans=n=getint(); m=getint(); for(int i=1,x,y;i<=m;i++){ x=getint(); y=getint(); g[x][y]=g[y][x]=1; du[x]++; du[y]++; } for(int i=1;i<=n;i++) ans-=du[i]&1; printf("%d\n",ans); for(int i=1;i<=n;i++) if(du[i]&1) while(du[i]) dfs(i); for(int i=1;i<=n;i++) while(du[i]) dfs(i); for(int i=1;i<=ls;i++) printf("%d %d\n",sa[i],sb[i]); ls=0; } }

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;  #define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 205, M = N * N * 20, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n, m;
int ind[N], oud[N];
int ST, ED;
int first[N]; int id;
int w[M], cap[M], nxt[M];
void ins(int x, int y, int cap_)
{  w[++id] = y;  cap[id] = cap_;  nxt[id] = first[x];  first[x] = id;  w[++id] = x;  cap[id] = 0;  nxt[id] = first[y];  first[y] = id;
}
int d[N];
bool bfs()
{  MS(d, -1);  queue<int>q; q.push(ST); d[ST] = 0;  while (!q.empty())  {  int x = q.front(); q.pop();  for (int z = first[x]; z; z = nxt[z])if (cap[z])  {  int y = w[z];  if (d[y] == -1)  {  d[y] = d[x] + 1;  q.push(y);  if (y == ED)return 1;  }  }  }  return 0;
}
int dfs(int x, int all)
{  if (x == ED)return all;  int use = 0;  for (int z = first[x]; z; z = nxt[z])if (cap[z])  {  int y = w[z];  if (d[y] == d[x] + 1)  {  int tmp = dfs(y, min(cap[z], all - use));  cap[z] -= tmp;  cap[z ^ 1] += tmp;  use += tmp;  if (use == all)break;  }  }  if (use == 0)d[x] = -1;  return use;
}
int dinic()
{  int ret = 0;  while (bfs())ret += dfs(ST, inf);  return ret;
}
int b[N], g;
void solve()
{  int sum = 0;  g = 0;  for (int i = 1; i <= n; ++i)  {  if (abs(ind[i] - oud[i]) % 2 == 1)b[++g] = i;  }  for (int i = 1; i <= g; i += 2)  {  ins(b[i], b[i + 1], 1);  ++oud[b[i]];  ++ind[b[i + 1]];  }  for (int i = 1; i <= n; ++i)  {  int w = abs(ind[i] - oud[i]) / 2;  if (ind[i] > oud[i])  {  ins(i, ED, w);  sum += w;  }  if (oud[i] > ind[i])  {  ins(ST, i, w);  }  }  dinic();  int ans = n - g;  printf("%d\n", ans);  for (int i = 2; i <= 2 * m; i += 2)  {  if (cap[i] == 0)printf("%d %d\n", w[i], w[i ^ 1]);  else printf("%d %d\n", w[i ^ 1], w[i]);  }
}
int main()
{  scanf("%d", &casenum);  for (casei = 1; casei <= casenum; ++casei)  {  scanf("%d%d", &n, &m);  ST = 0;  ED = n + 1;  MS(first, 0); id = 1;  for (int i = 1; i <= n; ++i)oud[i] = ind[i] = 0;  for (int i = 1; i <= m; ++i)  {  int x, y; scanf("%d%d", &x, &y);  ++oud[x];  ++ind[y];  ins(x, y, 1);  }  solve();  }  return 0;
}  

You are given an undirected connected graph consisting of n vertices and m edges. There are no loops and no multiple edges in the graph.

You are also given two distinct vertices s and t, and two values d_s and d_t. Your task is to build any spanning tree of the given graph (note that the graph is not weighted), such that the degree of the vertex s doesn't exceed d_s, and the degree of the vertex t doesn't exceed d_t, or determine, that there is no such spanning tree.

The spanning tree of the graph G is a subgraph which is a tree and contains all vertices of the graph G. In other words, it is a connected graph which contains n - 1 edges and can be obtained by removing some of the edges from G.

The degree of a vertex is the number of edges incident to this vertex.

If the answer doesn't exist print "No" (without quotes) in the only line of the output.

Otherwise, in the first line print "Yes" (without quotes). In the each of the next (n - 1) lines print two integers — the description of the edges of the spanning tree. Each of the edges of the spanning tree must be printed exactly once.

You can output edges in any order. You can output the ends of each edge in any order.

If there are several solutions, print any of them.