题目链接：点击打开链接
A. Contest time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output
Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully submitted it at the first attempt. Misha solved the problem that costs a points and Vasya solved the problem that costs b points. Besides, Misha submitted the problem c minutes after the contest started and Vasya submitted the problem d minutes after the contest started. As you know, on Codeforces the cost of a problem reduces as a round continues. That is, if you submit a problem that costs p points tminutes after the contest started, you get 
points.
Misha and Vasya are having an argument trying to find out who got more points. Help them to find out the truth.
Input
The first line contains four integers a, b, c, d (250 ≤ a, b ≤ 3500, 0 ≤ c, d ≤ 180).
It is guaranteed that numbers a and b are divisible by 250 (just like on any real Codeforces round).
Output
Output on a single line:
“Misha” (without the quotes), if Misha got more points than Vasya.
“Vasya” (without the quotes), if Vasya got more points than Misha.
“Tie” (without the quotes), if both of them got the same number of points.
Examples input
500 1000 20 30
output
Vasya
input
1000 1000 1 1
output
Tie
input
1500 1000 176 177
output
Misha
水题不多说~~
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int a,b,c,d;
int solve(int p,int t)
{
return max(3*p/10,p-p/250*t);
}
int main()
{
while(~scanf("%d %d %d %d",&a,&b,&c,&d))
{
int x=solve(a,c);
int y=solve(b,d);
if(x>y) puts("Misha");
else if(x<y) puts("Vasya");
else puts("Tie");
}
return 0;
} 题目链接：
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B. Misha and Changing Handles time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output
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.
Input
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 handlenew is not used and has not been used by anyone.
Output
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.
Examples input
5 Misha ILoveCodeforces Vasya Petrov Petrov VasyaPetrov123 ILoveCodeforces MikeMirzayanov Petya Ivanov
output
3 Petya Ivanov Misha MikeMirzayanov Vasya VasyaPetrov123
题解：（这道题用map也行得通）以下是字符串拷贝函数的使用，strcmp ( a，b ) 作用：是把字符串 b 连同串结束标志 ‘ \0 ‘ 复制到 a 中，也就是把串 a 换成 串 b
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n;
char a[50],b[50];
char str1[1010][50],str2[1010][50];
int main()
{
while(~scanf("%d",&n))
{
int num=0;
while(n--)
{
bool flag=0;
scanf("%s%s",a,b);
for(int i=0;i<num;i++)
{
if(strcmp(str2[i],a)==0)
{
strcpy(str2[i],b);
flag=1;
break;
}
}
if(!flag)
{
strcpy(str1[num],a);
strcpy(str2[num++],b);
}
}
printf("%d\n",num);
for(int i=0;i<num;i++)
printf("%s %s\n",str1[i],str2[i]);
}
return 0;
}
/* 第一种 map实现
#include<stdio.h>
#include<stdlib.h>
#include<cstring>
#include<iostream>
#include<map>
#include<algorithm>
using namespace std;
int main()
{
string a,b;
map<string,string>ma;
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
cin>>a>>b;
if(!ma.count(a)) // 指定元素出现的次数
{
ma[b]=a;
}
else
{
ma[b] = ma[a];
ma.erase(a);
}
}
printf("%d\n",ma.size()); // 返回 map中 元素的个数
map<string,string>::iterator it;
for(it=ma.begin();it!=ma.end();it++)//map函数的输出
cout<<it->second<<' '<<it->first<<endl;
return 0;
}*/
/* 第二种 map实现
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<map>
using namespace std;
int n;
string a,b;
map<string,string> M;
map<string,string>::iterator it;
int main()
{
scanf("%d",&n);
while(n--)
{
cin>>a>>b;
it=M.begin();
while(it!=M.end())
{
if(it->second==a)
break;
it++;
}
if(it!=M.end())
{
it->second=b;
}
else M[a]=b;
}
// printf("%d\n",M.size()); 这样写也行，但是 360当木马给截了，无语
int num=0;
for(it=M.begin();it!=M.end();it++)
num++;
printf("%d\n",num);
for(it=M.begin();it!=M.end();it++)
cout<<it->first<<" "<<it->second<<endl;
return 0;
}*/ 题目链接：
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C. Misha and Forest time limit per test 1 second memory limit per test 256 megabytes input standard input output 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, degreev and sv, 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 degreev and sv 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.
Input
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 degreei and si (0 ≤ degreei ≤ n - 1, 0 ≤ si < 216), separated by a space.
Output
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.
Examples input
3 2 3 1 0 1 0
output
2 1 0 2 0
input
2 1 1 1 0
output
1 0 1
Note
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”.
题意：给你一个无环无重边的图，并且给你每个点的与其相邻点的个数和相邻点标号的异或值，让你输出所有的边。
题解：其实给的图其实是一棵树，要解决此题就要从叶子节点入手，因为与叶子节点相邻的点只有一个，而叶子结点的异或值就是该点的标号，然后删去叶子结点，逐步往上递推就可以得到所有的边。
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<cstdlib>
#include<iostream>
using namespace std;
const int MAXN=(1<<16)+10;
int n;
int num[MAXN]; // num[i] = j 表示与 i相邻的点的个数为 j
int xsum[MAXN];// xsum[i] = j 表示与 j 相邻的点的异或和为 j
int ans[MAXN];
queue<int> Q;
int main()
{
while(~scanf("%d",&n))
{
pair<int,int> ans[MAXN];
while(!Q.empty()) Q.pop();
for(int i=0;i<n;i++)
{
scanf("%d%d",&num[i],&xsum[i]);
if(num[i]==1) Q.push(i); // 进队的是叶子结点
}
int cnt=0;
while(!Q.empty())
{
int from=Q.front(); // from是叶子结点
Q.pop();
if(num[from]==1) // ( num[]--的过程 from会为 0 )与叶子结点相邻的点的个数为 1
{
int to=xsum[from]; // to是叶子节点的父节点,因为叶子结点的异或和就是他的父节点
ans[cnt].first=from;
ans[cnt++].second=to; // 建立一条从叶子节点到父节点的边
num[to]--; // 删去叶子结点，故与父节点相邻的点的个数减 1
xsum[to]^=from; // 更新父节点的异或和。公式：a^b^c^a = b^c
if(num[to]==1)
Q.push(to);
}
}
printf("%d\n",cnt);
for(int i=0;i<cnt;i++)
printf("%d %d\n",ans[i].first,ans[i].second);
// printf("%d %d\n",ans[i].second,ans[i].first); 因为无向边，这样也可以
}
return 0;
}