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;
}