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*变元能取1、0两种状态，所以，最终答案是2^k，k表示*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;

}