uva 1397 - The Teacher's Side of Math(高斯消元)

题目链接：uva 1397 - The Teacher's Side of Math

题目大意：给出一个方程的解x=a1/m+b1/n,求原方程（给出系数即可）

解题思路：因为方程肯定等于0，所以对于各个系数ax/mby/n都会等于0，于是可以根据这个列出方程，注意最高项的系数始终为1.

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;
const int maxc = 20;
const int maxn = 40;

typedef long long ll;
typedef double Mat[maxn][maxn];

ll A, M, B, N, C[maxc+5][maxc+5];
Mat G;

void getc (int n) {
    for (int i = 0; i <= n; i++) {
        C[i][0] = C[i][i] = 1;
        for (int j = 1; j < i; j++)
            C[i][j] = C[i-1][j-1] + C[i-1][j];
    }
}

void init () {
    memset(G, 0, sizeof(G));
    int n = N * M;

    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= i; j++) {
            int l = j, r = i - j;
            double tmp = C[i][l] * pow(A * 1.0, l / M) * pow(B * 1.0, r / N);
            l %= M; r %= N;
            G[l*N+r][i] += tmp;
        }
    }
    G[n][n] = 1;
    G[n][n+1] = 1;
}

void gauss_elimin (int n) {
    /* for (int i = 0; i <= n; i++) { for (int j = 0; j <= n; j++) printf("%lf ", G[i][j]); printf("\n"); } */

    for (int i = 0; i < n; i++) {
        int r = i;

        for (int j = i + 1; j < n; j++) 
            if (fabs(G[j][i]) > fabs(G[r][i]))
                r = j;

        if (r != i) {
            for (int j = 0; j <= n; j++)
                swap(G[r][j], G[i][j]);
        }

        if (fabs(G[i][i]) < 1e-9)
            continue;

        for (int k = i + 1; k < n; k++) {
            double f = G[k][i] / G[i][i];
            for (int j = 0; j <= n; j++)
                G[k][j] -= G[i][j] * f;
        }
    }

    for (int i = n-1; i >= 0; i--) {
        for (int j = i+1; j < n; j++)
            G[i][n] -= G[j][j] * G[i][j];
        G[i][i] = G[i][n] / G[i][i];
        if (fabs(G[i][i]) < 1e-9)
            G[i][i] = 0;
    }

    printf("%.0f", G[n-1][n-1]);
    for (int i = n-2; i >= 0; i--)
        printf(" %.0f", G[i][i]);
    printf("\n");
}

int main () {
    getc(maxc);

    while (scanf("%lld%lld%lld%lld", &A, &M, &B, &N) == 4 && A + M + B + N) {
        init();
        gauss_elimin (N * M + 1);
    }
    return 0;
}