HDU 5755 Gambler Bo 高斯消元解同余方程组

来源：互联网发布：网络工商营业执照编辑：程序博客网时间：2024/05/20 18:54

参考：http://www.cnblogs.com/yohaha/p/5710264.html

2016 Multi-University Training Contest 3 1004 HDU 5755 Gambler Bo

高斯消元解同余方程组，我做的高斯消元的第一个题，解法完全是参考别人的，写博客就是存个板。

#include<stdio.h>#include<algorithm>#include<iostream>#include<string.h>#include<math.h>using namespace std;const int MAXN=905;int dx[] = {0,1,0,-1};int dy[] = {1,0,-1,0};int b[MAXN];int a[MAXN][MAXN];//增广矩阵int x[MAXN];//解集bool free_x[MAXN];//标记是否是不确定的变元int n,m;const int mod = 3;int gcd(int a, int b){    return b?gcd(b, a%b):a;}int lcm(int a, int b){    return a/gcd(a, b)*b;}long long inv(long long a, long long m){    if(a == 1)        return 1;    return inv(m%a, m)*(m-m/a)%m;}void gauss(int equ, int var){    int max_r, col, k;    for(k = 0, col = 0; k < equ && col < var; k++, col++)    {        max_r = k;        for(int i = k + 1; i < equ; i ++)        {            if(abs(a[i][col]) > abs(a[max_r][col]))                max_r = i;        }        if(a[max_r][col] == 0)        {            k--;            continue;        }        if(max_r != k)        {            for(int j = col; j < var+1; j++)            {                swap(a[k][j], a[max_r][j]);            }        }        for(int i = k + 1; i < equ; i++)        {            if(a[i][col])            {                int LCM = lcm(abs(a[i][col]), abs(a[k][col]));                int ta = LCM/abs(a[i][col]);                int tb = LCM/abs(a[k][col]);                if(a[i][col] * a[k][col] < 0)                    tb = -tb;                for(int j = col; j < var+1; j++)                {                    a[i][j] = ((a[i][j]*ta - a[k][j]*tb)%mod+mod)%mod;                }            }        }        for(int i = var-1; i >= 0; i--)        {            int tmp = a[i][var];            for(int j = i+1; j < var; j++)            {                if(a[i][j])                {                    tmp -= a[i][j]*x[j];                    tmp = (tmp%mod+mod)%mod;                }            }            x[i] = a[i][i]*tmp%mod;        }    }}bool ok(int x){    if(x >= 0 && x < m*n)        return 1;    return 0;}int main(void){    int t;    scanf("%d", &t);    while(t--)    {        memset(x,0,sizeof(x));        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        scanf("%d%d", &m, &n);        for(int i = 0; i < m * n; ++i)            scanf("%d", &b[i]);        for(int i = 0; i < m * n; ++i)        {            int r = i + 1, l = i - 1, u = i - n, d = i + n;            if(ok(r) && (i+1) % n != 0)                a[i][r] = 1;            if(ok(l) && i % n != 0)                a[i][l] = 1;            if(ok(u))                a[i][u] = 1;            if(ok(d))                a[i][d] = 1;            a[i][i] = 2;            a[i][m*n] = (3 - b[i]) % 3;        }        gauss(n*m,n*m);        int cnt = 0;        for(int i = 0; i < n*m; ++i)            cnt += x[i];        printf("%d\n", cnt);        for(int i = 0; i < n*m; ++i)            while(x[i])            {                printf("%d %d\n", i/n+1,i%n+1);                x[i]--;            }    }return 0;}

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