HDU 1850 Being a Good Boy in Spring Festival 基础Nim博弈

题目大意:

就是Nim游戏从SG != 0的点向 = 0的点有几种走法


大致思路:

如果想改变第 i 堆使得SG = 0, 那么第 i 对要变成其他堆的SG和的数量, 于是判断一下大小就可以了


代码如下:

Result  :  Accepted     Memory  :  1576 KB     Time  :  0 ms

/*
 * Author: Gatevin
 * Created Time:  2015/5/6 20:18:10
 * File Name: Rin_Tohsaka.cpp
 */
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;
#define foreach(e, x) for(__typeof(x.begin()) e = x.begin(); e != x.end(); ++e)
#define SHOW_MEMORY(x) cout<<sizeof(x)/(1024*1024.)<<"MB"<<endl

int m;
int num[110];

int main()
{
    while(scanf("%d", &m), m)
    {
        int sg = 0;
        for(int i = 1; i <= m; i++)
            scanf("%d", num + i), sg ^= num[i];
        int ans = 0;
        if(sg == 0) puts("0");
        else
        {
            for(int i = 1; i <= m; i++)
                if((sg^num[i]) < num[i]) ans++;
            printf("%d\n", ans);
        }
    }
    return 0;
}


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