Hdu 5229 ZCC loves strings（简单题）

题目链接

ZCC loves strings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 343    Accepted Submission(s): 131

Problem Description

ZCC has got N strings. He is now playing a game with Miss G.. ZCC will pick up two strings among those N strings randomly(A string can't be chosen twice). Each string has the same probability to be chosen. Then ZCC and Miss G. play in turns. Miss G. always plays first. In each turn, the player can choose operation A or B.
  
Operation A: choose a non-empty string between two strings, and delete a single letter at the end of the string.
    
Operation B: When two strings are the same and not empty, empty both two strings.
  
The player who can't choose a valid operation loses the game.
  
ZCC wants to know what the probability of losing the game(i.e. Miss G. wins the game) is.

 

Input

The first line contains an integer T(T≤5) which denotes the number of test cases.
  
For each test case, there is an integer N(2≤N≤20000) in the first line. In the next N lines, there is a single string which only contains lowercase letters. It's guaranteed that the total length of strings will not exceed 200000.

 

Output

For each test case, output an irreducible fraction "p/q" which is the answer. If the answer equals to 1, output "1/1" while output "0/1" when the answer is 0.

 

Sample Input

   
   
   
   

    
    
    
    1
3
xllendone
xllendthree
xllendfour
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    2/3
   
   
   
   

题意：有n个字符串，从这里面随机选择两个字符串。然后两个人轮流操作，有两种操作：

1，如果两个字符串相同，可以删除这两个字符串。

2，选择一个字符串，删掉最后一个字符。

第一个不能操作的人输。

对于先手来说：

如果两个串相同，则先手赢

否则，由于两个人都采用最优策略，则两个串相同的情况一定不会出现，那么如果两个串的总共有奇数个字符则先手赢，否则先手输。

那么我们只需要统计奇数串和偶数串的个数，以及相同的串的个数就行了。

代码如下：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<stdlib.h>
#include<vector>
#define inff 0x3fffffff
#define nn 21000
#define mod 1000000007
typedef __int64 LL;
typedef unsigned __int64 LLU;
const LL inf64=inff*(LL)inff;
using namespace std;
int n;
map<string,int>ma;
char s[nn*10];
int gcd(int x,int y)
{
    if(y==0)
    {
        return x;
    }
    return gcd(y,x%y);
}
int main()
{
    int t,i;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        ma.clear();
        int ix,fc;
        ix=fc=0;
        int ans=0;
        for(i=1;i<=n;i++)
        {
            scanf("%s",s);
            int ls=strlen(s);
            if(ls%2)
                ix++;
            else
                fc++;
            if(ma.count(s)==0)
            {
                ma[s]=1;
            }
            else
            {
                ans+=ma[s];
                ma[s]++;
            }
        }
        ans+=ix*fc;
        if(ans==0)
        {
            puts("0/1");
            continue;
        }
        int tem=gcd(ans,(n-1)*n/2);
        printf("%d/%d\n",ans/tem,(n-1)*n/2/tem);
    }
    return 0;
}