线性方程组的追赶式直接解法

#include <stdio.h> const int maxn = 15; int main() { /* 数据存储格式: a[]对角线上, b[]对角线, c[]对角线下, f[]右端常数 double a[N] = {0, 0, -1, -1, -1, -1, -1, -1, -1, -1, -1}; double b[N] = {0, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4}; double c[N] = {0, -1, -1, -1, -1, -1, -1, -1, -1, -1, 0}; double f[N] = {7, 5, -13, 2, 6, -12, 14, -4, 5, -5}; double x[N] = {0}, u[N] = {0}, y[N] = {0}; */ freopen("zgf.txt", "r", stdin); freopen("ans.txt", "w", stdout); double a[maxn] = {0}, b[maxn] = {0}, c[maxn] = {0}, f[maxn] = {0}, x[maxn] = {0}, u[maxn] = {0}, y[maxn] = {0}; int n, N; scanf("%d", &n); N = n + 1; //0列不使用 scanf("%lf %lf %lf", &b[1], &a[2], &f[1]); for(int i = 2; i < N - 1; i++) scanf("%lf %lf %lf %lf", &c[i - 1], &b[i], &a[i + 1], &f[i]); scanf("%lf %lf %lf", &c[N - 1], &b[N - 1], &f[N - 1]); //追 u[1] = c[1] / b[1]; //L1 == b[1] y[1] = f[1] / b[1]; for(int i = 2; i < N - 1; i++) { double l = b[i] - a[i] * u[i - 1]; //克洛特分解A = LU u[i] = c[i] / l; y[i] = (f[i] - a[i] * y[i - 1]) / l; } y[N - 1] = (f[N - 1] - a[N - 1] * y[N - 2]) / (b[N - 1] - a[N - 1] * u[N - 2]); //赶 x[N - 1] = y[N - 1]; for(int i = N - 2; i >= 1; i--) x[i] = y[i] - u[i] * x[i + 1]; for(int i = 1; i < N; i++) printf("x%d = %f\n", i, x[i]); return 0; }