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追赶法_解线性方程组的直接解法

标签：计算方法实验

#include 
   
    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;}
   

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