UVA 1397 - The Teacher's Side of Math(高斯消元)
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UVA 1397 - The Teacher‘s Side of Math
题意:给定一个x=a1/m+b1/n。求原方程组
思路:因为m*n最多20,全部最高项仅仅有20。然后能够把每一个此项拆分。之后得到n种不同无理数,每一项为0。就能够设系数为变元。构造方程进行高斯消元
一開始用longlong爆了。换成分数写法也爆了,又不想改高精度。最后是机智的用了double型过的,只是用double精度问题,所以高斯消元的姿势要正确,而且最后输出要注意-0的情况
代码:
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; typedef long long ll; const int N = 25; const double eps = 1e-9; ll a, m, b, n, C[N][N]; int hash[N][N], tot; double A[N][N]; void build() { memset(A, 0, sizeof(A)); A[0][0] = 1; for (int i = 1; i <= tot; 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; A[hash[l][r]][i] += tmp; } } A[tot][tot] = 1; A[tot][tot + 1] = 1; tot++; } void getC() { for (int i = 0; i <= 20; 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 gethash() { tot = 0; for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { hash[i][j] = tot++; } } } void print(double x) { char s[100]; sprintf(s, "%.0lf", x); if (strcmp(s, "-0") == 0) printf(" %s", s + 1); else printf(" %s", s); } void gauss() { for (int i = 0; i < tot; i++) { int r = i; for (int j = i + 1; j < tot; j++) { if (fabs(A[j][i]) > fabs(A[r][i])) r = j; } if (fabs(A[r][i]) < eps) continue; for (int j = i; j <= tot; j++) swap(A[r][j], A[i][j]); for (int j = 0; j < tot; j++) { if (i == j) continue; if (fabs(A[j][i]) >= eps) { double tmp = A[j][i] / A[i][i]; for (int k = i; k <= tot; k++) A[j][k] -= tmp * A[i][k]; } } } printf("1"); for (int i = tot - 2; i >= 0; i--) print(A[i][tot] / A[i][i]); printf("\n"); } int main() { getC(); while (~scanf("%lld%lld%lld%lld", &a, &m, &b, &n)) { if (!a && !m && !b && !n) break; gethash(); build(); gauss(); } return 0; }
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