POJ1830（异或高斯消元）

对于某个开关，都有n个选项可能影响它的结果，如果会影响，则系数为1，否则系数为0；最后得到自由元的个数，自由元可选0也可选1.

#include <cstdio>
#include <algorithm>

int T, n, a[30], x, y;

int gauss() {
    for (int i = 1; i <= n; i++) {
        //列主
        for (int j = i + 1; j <= n; j++) {
            if (a[j] > a[i]) {
                std::swap(a[i], a[j]);
            }
        }
        if (a[i] == 0)  return 1 << (n - i + 1);
        if (a[i] == 1)  return -1;
        //消元
        for (int k = n; k; k--) {
            if (a[i] & (1 << k)) {
                for (int j = 1; j <= n; j++) {
                    if (i != j && a[j] & (1 << k)) {
                        a[j] ^= a[i];
                    }
                }
                break;
            }
        }
    }
    return 1;
}

int main(int argc, char const *argv[]) {
    scanf("%d", &T);
    while (T--) {
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) {
            scanf("%d", &a[i]); 
        }
        for (int i = 1, j; i <= n; i++) {
            scanf("%d", &j);
            a[i] ^= j;//等号右侧
            a[i] |= 1 << i;//a[i][i] = 1
        }
        while (~scanf("%d %d", &x, &y) && (x | y)) {
            a[y] |= 1 << x;//a[y][x] = 1
        }
        int ans = gauss();
        if (ans > 0)    printf("%d\n", ans);
        else    puts("Oh,it's impossible~!!");
    }
    return 0;
}