HDU5229-ZCC loves strings

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本文介绍了一个关于字符串的游戏，玩家通过删除字符串中的字符进行博弈，分析了先手玩家获胜的概率，并给出了具体的算法实现。

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ZCC loves strings

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 939 Accepted Submission(s): 362

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

Source

BestCoder Round #41

Recommend

hujie

题意：有N个字符串，从这N个串中等概率随机选两个字符串（不可以是同一个）。然后两个人轮流操作，有两种操作：如果两个字符串相同，可以删除这两个字符串；不相同则选择一个字符串，删掉最后一个字符。第一个不能操作的人输。问先手获胜的概率

解题思路：对于先手来说，如果两个串相同，则先手赢。如果两个串的总共有奇数个字符则先手赢，否则先手输。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>

using namespace std;

string s[20009];

int gcd(int x,int y)
{
    if(x>y) return gcd(y,x);
    while(y%x)
    {
        int t=x;
        x=y%x;
        y=t;
    }
    return x;
}

int main()
{
    int t,n,a,b;
    scanf("%d",&t);
    while(t--)
    {
        a=b=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            cin>>s[i];
            if(s[i].length()%2) a++;
            else b++;
        }
        sort(s,s+n);
        int sum=n*(n-1)/2;
        int sum1=a*b;
        int k=1;
        for(int i=1;i<n;i++)
        {
            if(s[i]==s[i-1]) k++;
            else
            {
                sum1+=(k-1)*k/2;
                k=1;
            }
        }
        sum1+=(k-1)*k/2;
        printf("%d/%d\n",sum1/gcd(sum1,sum),sum/gcd(sum,sum1));
    }
    return 0;
}