/******************************

 code by drizzle

 blog: www.cnblogs.com/hsd-/

 ^ ^    ^ ^

  O      O

 ******************************/

 #include<bits/stdc++.h>

 #include<iostream>

 #include<cstring>

 #include<cmath>

 #include<cstdio>

 #define ll long long

 #define mod 1000000007

 #define PI acos(-1.0)

 #define N 1000000000

 using namespace std;

 int a,b,c,d;

 int main()

 {

     scanf("%d %d %d %d",&a,&b,&c,&d);

     int s1=max(a/*,a-a/*c);

     int s2=max(b/*,b-b/*d);

     if(s1==s2)

     {

         cout<<"Tie"<<endl;

     }

     else

     {

         if(s1>s2)

         {

             cout<<"Misha"<<endl;

         }

         else

             cout<<"Vasya"<<endl;

     }

     return ;

 }

Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.

Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.

The first line contains integer q (1 ≤ q ≤ 1000), the number of handle change requests.

Next q lines contain the descriptions of the requests, one per line.

Each query consists of two non-empty strings old and new, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings old and new are distinct. The lengths of the strings do not exceed 20.

The requests are given chronologically. In other words, by the moment of a query there is a single person with handle old, and handle newis not used and has not been used by anyone.

In the first line output the integer n — the number of users that changed their handles at least once.

In the next n lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, old andnew, separated by a space, meaning that before the user had handle old, and after all the requests are completed, his handle is new. You may output lines in any order.

Each user who changes the handle must occur exactly once in this description.

 /******************************

 code by drizzle

 blog: www.cnblogs.com/hsd-/

 ^ ^    ^ ^

  O      O

 ******************************/

 #include<bits/stdc++.h>

 #include<iostream>

 #include<cstring>

 #include<cmath>

 #include<cstdio>

 #define ll long long

 #define mod 1000000007

 #define PI acos(-1.0)

 #define N 1000000000

 using namespace std;

 map<int ,string> mp1;

 map<string,int> mp2;

 map<string,string> mp3;

 map<string,string> mp4;

 string str1,str2;

 int n;

 int main()

 {

     scanf("%d",&n);

     int jishu=;

     for(int i=;i<=n;i++)

     {

         cin>>str1>>str2;

         if(mp2[str1]==)

         {

             mp1[jishu++]=str1;

             mp2[str1]=;

             mp2[str2]=;

             mp3[str1]=str2;

             mp4[str2]=str1;

         }

         else

         {

             mp3[mp4[str1]]=str2;

             mp4[str2]=mp4[str1];

             mp2[str1]=;

             mp2[str2]=;

         }

     }

     printf("%d\n",jishu);

     for(int i=;i<jishu;i++)

         cout<<mp1[i]<<" "<<mp3[mp1[i]]<<endl;

     return ;

 }

1 second

256 megabytes

standard input

standard output

Let's define a forest as a non-directed acyclic graph (also without loops and parallel edges). One day Misha played with the forest consisting of n vertices. For each vertex v from 0 to n - 1 he wrote down two integers, degree_v and s_v, were the first integer is the number of vertices adjacent to vertex v, and the second integer is the XOR sum of the numbers of vertices adjacent to v (if there were no adjacent vertices, he wrote down 0).

Next day Misha couldn't remember what graph he initially had. Misha has values degree_v and s_v left, though. Help him find the number of edges and the edges of the initial graph. It is guaranteed that there exists a forest that corresponds to the numbers written by Misha.

The first line contains integer n (1 ≤ n ≤ 2¹⁶), the number of vertices in the graph.

The i-th of the next lines contains numbers degree_i and s_i (0 ≤ degree_i ≤ n - 1, 0 ≤ s_i < 2¹⁶), separated by a space.

In the first line print number m, the number of edges of the graph.

Next print m lines, each containing two distinct numbers, a and b (0 ≤ a ≤ n - 1, 0 ≤ b ≤ n - 1), corresponding to edge (a, b).

Edges can be printed in any order; vertices of the edge can also be printed in any order.

The XOR sum of numbers is the result of bitwise adding numbers modulo 2. This operation exists in many modern programming languages. For example, in languages C++, Java and Python it is represented as "^", and in Pascal — as "xor".

 /******************************

 code by drizzle

 blog: www.cnblogs.com/hsd-/

 ^ ^    ^ ^

  O      O

 ******************************/

 #include<bits/stdc++.h>

 #include<iostream>

 #include<cstring>

 #include<cmath>

 #include<cstdio>

 #define ll long long

 #define mod 1000000007

 #define PI acos(-1.0)

 using namespace std;

 struct node

 {

     int pos;

     int num;

     int sum;

     /*friend bool operator < (node aa,node bb)

     {

         return aa.num<bb.num;

     }

     */

 } N[];

 struct ans

 {

     int x,y;

 } NN[];

 queue<node> p;

 int n,vis[];

 int main()

 {

     scanf("%d",&n);

     for(int i=; i<n; i++)

     {

         N[i].pos=i;

         scanf("%d %d",&N[i].num,&N[i].sum);

         if(N[i].num==)

             p.push(N[i]);

     }

     int jishu=;

     while(!p.empty())

     {

         node exm=p.front();

         p.pop();

         if(exm.num ==  || vis[exm.pos]) continue;

         NN[++jishu].x=exm.sum;

         NN[jishu].y=exm.pos;

         N[exm.sum].num--;

         if(N[exm.sum].num == ) vis[exm.sum] = ;

         N[exm.sum].sum^=exm.pos;

         if(N[exm.sum].num==)

             p.push(N[exm.sum]);

     }

     printf("%d\n",jishu);

     for(int i=; i<=jishu; i++)

         printf("%d %d\n",NN[i].x,NN[i].y);

     return ;

 }