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位运算。。。。

Revenge of Nim II

3
1
2
3
2 2 2
5
1 2 3 4 5

No
Yes
Yes
Hint
For the third test case, the second player can move heaps with 4 and 5 objects out, so the nim-sum of the sizes of the left heaps is 1⊕2⊕3 = 0.
 

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

using namespace std;

typedef long long int LL;

int n;
LL c[60];
LL a[1100];

int main()
{
    int T_T;
    scanf("%d",&T_T);
    while(T_T--)
    {
        memset(c,0,sizeof(c));
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            cin>>a[i];
        bool flag=false;
        for(int i=1;i<=n;i++)
        {
            for(int j=60;j>=0;j--)
            {
                if(c[j]==0&&(a[i]&(1LL<<j)))
                {
                    c[j]=a[i];
                    break;
                }
                else if(a[i]&(1LL<<j))
                {
                    a[i]^=c[j];
                    if(a[i]==0) flag=true;
                }
            }
        }
        if(flag==true) puts("Yes");
        else puts("No");
    }
    return 0;
}

HDOJ 5088 Revenge of Nim II 位运算

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原文地址：http://blog.csdn.net/ck_boss/article/details/40707743