kiki's game

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 40000/10000 K (Java/Others)
Total Submission(s): 11452 Accepted Submission(s): 6944

Problem Description

Recently kiki has nothing to do. While she is bored, an idea appears in his mind, she just playes the checkerboard game.The size of the chesserboard is n*m.First of all, a coin is placed in the top right corner(1,m). Each time one people can move the coin into the left, the underneath or the left-underneath blank space.The person who can't make a move will lose the game. kiki plays it with ZZ.The game always starts with kiki. If both play perfectly, who will win the game?

Input

Input contains multiple test cases. Each line contains two integer n, m (0<n,m<=2000). The input is terminated when n=0 and m=0.

Output

If kiki wins the game printf "Wonderful!", else "What a pity!".

Sample Input

5 3 5 4 6 6 0 0

Sample Output

What a pity! Wonderful! Wonderful!

Author

月野兔

Source

HDU 2007-11 Programming Contest

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已知Nim博弈的必败态条件是：所有集合的异或是0.

这个题目要求有多少种操作，就是有有多少种方式可以由当前状态转移到一个必败态

首先求出异或和，判断当前状态

如果是必胜态，那么尝试着从其中一个集合取数，让这个集合和其他集合的异或和为0.

其他集合的异或和可用 sum^a[i]来求出 sum^a[i]就等价于a[]中除了a[i]的异或和，因为异或是可逆的操作

然后判断当前的a[i]如果大于sum^a[i]，就可以操作这个数字，ans++

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<deque>
#include<iomanip>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<functional>
#include<fstream>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;

#define N 100
#define MAXN 20000 + 9
#define INF 1000000009
#define eps 0.00000001
#define sf(a) scanf("%d",&a)

int n;
int a[N];
int main()
{
    while (sf(n), n)
    {
        int sum = 0;
        for (int i = 0; i < n; i++)
            sf(a[i]), sum ^= a[i];
        if (sum == 0)
            cout << 0 << endl;
        else
        {
            int ans = 0;
            for (int i = 0; i < n; i++)
            {
                if ((sum^a[i]) < a[i])
                    ans++;
            }
            cout << ans << endl;
        }
    }
}