hdu5755 Gambler Bo 题解 高斯消元解模线性方程组

题目链接：https://acm.hdu.edu.cn/showproblem.php?pid=5755

题目大意：略

解题思路：略（因为是模板题）

示例程序：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 33 * 33;

int T, N, M, n, b[33][33], a[maxn][maxn];

int id(int x, int y) {
    return (x - 1) * M + y;
}

void gauss() {
    for (int i = 1; i <= n; i++) {
        int r = i;
        for (int j = i; j <= n; j++) {
            if (a[j][i]) {
                r = j;
                break;
            }
        }
        if (!a[r][i]) {
            if (a[r][n+1]) assert(1 == 0); // 应该是不会出现的
            else continue;
        }
        for (int j = i; j <= n+1; j++)
            swap(a[r][j], a[i][j]);
        for (int j = 1; j <= n; j++) {
            if (j == i || !a[j][i])
                continue;
            int tmp = (a[j][i] == a[i][i]) ? 1 : 2;
            for (int k = 1; k <= n+1; k++) {
                a[j][k] = (a[j][k] * tmp - a[i][k] + 3) % 3;
            }
        }
    }
}

int main() {
    scanf("%d", &T);
    while (T--) {
        scanf("%d%d", &N, &M);
        n = N * M;
        for (int i = 1; i <= N; i++)
            for (int j = 1; j <= M; j++)
                scanf("%d", &b[i][j]);
        memset(a, 0, sizeof a);
        for (int x = 1, i = 1; x <= N; x++)
            for (int y = 1; y <= M; y++, i++)
                a[i][n+1] = (3 - b[x][y]) % 3;
        for (int i = 1; i <= N; i++) {
            for (int j = 1; j <= M; j++) {
                a[id(i, j)][id(i, j)] = 2;
                if (j < M) a[id(i, j+1)][id(i, j)] = 1;
                if (j > 1) a[id(i, j-1)][id(i, j)] = 1;
                if (i < N) a[id(i+1, j)][id(i, j)] = 1;
                if (i > 1) a[id(i-1, j)][id(i, j)] = 1;
            }
        }
        gauss();
        int ans = 0;
        for (int i = 1; i <= n; i++)
                if (a[i][n+1])
                    ans += a[i][n+1] * a[i][i] % 3;
        printf("%d\n", ans);
        for (int x = 1, i = 1; x <= N; x++)
            for (int y = 1; y <= M; y++, i++)
                if (a[i][n+1]) {
                    int cnt = a[i][n+1] * a[i][i] % 3;
                    while (cnt--)
                    printf("%d %d\n", x, y);
                }
    }
    return 0;
}