HDU - 5088： Revenge of Nim II （问是否存在子集的异或为0）

标签：before   lin   mes   course   The   nic   ==   存在   names   

InputThe first line contains a single integer T, indicating the number of test cases. 

Each test case begins with an integer N, indicating the number of heaps. Then N integer Ai follows, indicating the number of each heap. 

[Technical Specification] 
1. 1 <= T <= 100 
2. 1 <= N <= 1 000 
3. 1 <= Ai <= 1 000 000 000 000OutputFor each test case, output “Yes” if the second player can win by moving some (can be zero) heaps out, otherwise “No”.Sample Input

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

Sample Output

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.
        

把上一题的S改为0即可。依然占位。

#include<bits/stdc++.h>
#define lowbit(x) (x&-x)
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=10100;
ll a[maxn],x1[maxn],x2[maxn],c[65];
int main()
{
    int T,N,cnt;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&N);
        rep(i,1,N) scanf("%lld",&a[i]);
        if(N==1) puts("No");
        else {
            ll S=0; cnt=0;
            rep(i,0,42) c[i]=0;
            rep(i,1,N){
              rep(j,1,cnt){
                if(a[i]&lowbit(c[j])) a[i]^=c[j];
             }if(a[i]!=0) c[++cnt]=a[i];
            }
            rep(i,1,cnt) if(S&lowbit(c[i])) S^=c[i];
            if(S!=0||cnt==N) puts("No");
            else puts("Yes");
        }
    }
    return 0;
}

HDU - 5088： Revenge of Nim II （问是否存在子集的异或为0）

标签：before   lin   mes   course   The   nic   ==   存在   names   

原文地址：https://www.cnblogs.com/hua-dong/p/9735927.html