http://acm.hdu.edu.cn/showproblem.php?pid=5088

求能否去掉几堆石子使得nim游戏胜利

我们可以把题目转化成求n堆石子中的k堆石子数异或为0的情况数。使用x1---xn表示最终第i堆石子到底取不取（1取，0不取），将每堆石子数画成2进制的形式，列成31个方程来求自由变元数，最后由于自由变元能取1、0两种状态，所以自由变元数多于0即可输出Yes。

注意有40+个方程，因为A[I]<=1e12....

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define clr0(x) memset(x,0,sizeof(x))
#define clr1(x) memset(x,-1,sizeof(x))
#define eps 1e-9
const double pi = acos(-1.0);
typedef long long LL;
typedef unsigned long long ULL;
const int modo = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int inf = 0x3fffffff;
const LL _inf = 1e18;
const int maxn = 1005,maxm = 10005;

#define MAXN 1100
#define MOD 1000007
#define weishu 42
LL a[MAXN], g[MAXN][MAXN];
int Gauss(int n) {
    int i, j, r, c, cnt;
    for (c = cnt = 0; c < n; c++) {
        for (r = cnt; r < weishu; r++) {
            if (g[r][c])
                break;
        }
        if (r < weishu) {
            if (r != cnt) {
                for (i = 0; i < n; i++)
                    swap(g[r][i], g[cnt][i]);
            }
            for (i = cnt + 1; i < weishu; i++) {
                if (g[i][c]) {
                    for (j = 0; j < n; j++)
                        g[i][j] ^= g[cnt][j];
                }
            }
            cnt++;
        }
    }
    return n - cnt;
}
int main() {
    int c;
    int n, i, j;
    int ans, vary;
    scanf("%d", &c);
    while (c--) {
        int fuck = 0;
        scanf("%d", &n);
        for (i = 0; i < n; i++){
            scanf("%I64d", &a[i]);
            fuck ^= a[i];
        }
        for (i = 0; i < weishu; i++) {
            for (j = 0; j < n; j++)
                g[i][j] = (a[j] >> i) & 1;
        }
        vary = Gauss(n);
        //printf("%d\n", vary);
        if(vary <= 0 )//|| (fuck == 0 && vary <= 1))
            puts("No");
        else
            puts("Yes");

    }
    return 0;
}

hdu 5088 高斯消元n堆石子取k堆石子使剩余异或值为0