http://acm.hdu.edu.cn/showproblem.php?pid=3915

这道题目是和博弈论挂钩的高斯消元。本题涉及的博弈是nim博弈，结论是：当先手处于奇异局势时（几堆石子数相互异或为0），其必败。

思路在这里，最后由于自由变元能取1、0两种状态，所以，最终答案是2^k，k表示自由变元的个数。

#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 110
#define MOD 1000007
#define weishu 31
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);
        LL ans = 1;
        while(vary--){
            ans <<= 1;
            ans %= MOD;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}

hdu 3915 高斯消元