hdu 5755(高斯消元)-优快云博客

本文介绍了一种使用高斯消元法解决模方程组的算法实现，通过矩阵运算来求解特定形式的模方程组，并展示了如何利用该算法进行求解的具体步骤。

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列出模方程,高斯消元解

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;


#define N 1005
#define M 400200
#define U 1000000
#define inf 0x3f3f3f3f
#define mod 3
#define LL long long
#define ULL unsigned long long
#define ls (i << 1)
#define rs (ls | 1)
#define md (ll + rr >> 1)
#define lson ll, md, ls
#define rson md + 1, rr, rs
#define MP make_pair

int gcd(int a, int b){
    return b == 0 ? a : gcd(b, a % b);
}
int lcm(int a, int b){
    return a * b / gcd(a, b);
}
int a[N][N], x[N], v[N][N];
vector<int> ans;
int d[4][2] = {-1, 0, 0, 1, 1, 0, 0, -1};
void Gauss(int n){
    int k, col;
    for(k = 0, col = 0; k < n && col < n; ++k, ++col){
        int mx = k;
        for(int i = k + 1; i < n; ++i)
            if(abs(a[i][col]) > abs(a[mx][col]))
               mx = i;
        if(k != mx){
            for(int i = col; i < n; ++i)
                swap(a[k][i], a[mx][i]);
            swap(x[k], x[mx]);
        }
        if(a[k][col] == 0){
            k--; continue;
        }
        for(int i = k + 1; i < n; ++i){
            if(a[i][col]){
                int LCM = lcm(a[i][col], a[k][col]);
                int t1 = LCM / a[i][col], t2 = LCM / a[k][col];
                for(int j = col; j < n; ++j)
                    a[i][j] = ((a[i][j] * t1 - a[k][j] * t2) % mod + mod) % mod;
                x[i] = ((x[i] * t1 - x[k] * t2) % mod + mod) % mod;
            }
        }
    }
    for(int i = k - 1; i >= 0; --i){
        for(int j = i + 1; j < n; ++j)
            x[i] = ((x[i] - x[j] * a[i][j]) % mod + mod) % mod;
        x[i] = x[i] * a[i][i] % mod;
        if(x[i] < 0) x[i] += mod;
    }
}
int n, m;
bool check(int x, int y){
    return x >= 0 && x < n && y >= 0 && y < m;
}
int main(){
    int cas;
    scanf("%d", &cas);
    while(cas--){
        scanf("%d%d", &n, &m);
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < m; ++j)
                scanf("%d", &v[i][j]);
        memset(x, 0, sizeof x);
        memset(a, 0, sizeof a);
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < m; ++j){
                int cur = i * m + j;
                a[cur][cur] = 2;
                for(int k = 0; k < 4; ++k){
                    int nx = i + d[k][0], ny = j + d[k][1];
                    if(check(nx, ny)){
                        int nxt = nx * m + ny;
                        a[cur][nxt] = 1;
                    }
                }
                x[cur] = (3 - v[i][j]) % 3;
            }
        Gauss(n * m);
        ans.clear();
        for(int i = 0; i < n * m; ++i){
            while(x[i]){
                ans.push_back(i);
                x[i]--;
            }
        }
        printf("%d\n", (int)ans.size());
        for(int i = 0; i < ans.size(); ++i){
            int r = ans[i] / m;
            int c = ans[i] % m;
            printf("%d %d\n", r + 1, c + 1);
        }
    }
    return 0;
}