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

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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;}

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