ACM学习历程—HDU 3915 Game（Nim博弈 && xor高斯消元）

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <string>
#define LL long long

using namespace std;

const int maxN = 105;
const int MOD = 1000007;
int n, s[maxN];

void input()
{
    scanf("%d", &n);
    for (int i = 0; i < n; ++i)
        scanf("%d", &s[i]);
}

//xor高斯消元求线性基
//时间复杂度O(30n)
int xorGauss(int n)
{
    int row = 0;
    for (int i = 30; i >= 0; i--)
    {
        int j;
        for (j = row; j < n; j++)
            if(s[j]&(1<<i))
                break;
        if (j != n)
        {
            swap(s[row], s[j]);
            for (j = 0; j < n; j++)
            {
                if(j == row) continue;
                if(s[j]&(1<<i))
                    s[j] ^= s[row];
            }
            row++;
        }
    }
    return row;
}

void work()
{
    int row, ans, k;
    row = xorGauss(n);
    ans = n-row;
    if (ans != -1)
    {
        k = 1;
        while (ans)
        {
            k <<= 1;
            k %= MOD;
            ans--;
        }
        ans = k;
    }
    else
        ans = -1;
    printf("%d\n", ans);
}

int main()
{
    //freopen("test.in", "r", stdin);
    int T;
    scanf("%d", &T);
    for (int times = 0; times < T; ++times)
    {
        input();
        work();
    }
    return 0;
}