题意:给定一个一元多次方程组,要求求出所有根
思路:利用牛顿迭代法 xn+1=xn?f(xn)/f′(xn),不断迭代就能求出较为精确的值,然后由于有的方程可能有多解,每次解得一个X后,就把原式子除以(x - X),这个是肯定能整除的,把方程降阶然后继续用牛顿迭代法直到求出所有解
代码:
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; const int N = 10; int n; double a[N]; double cal(double *f, double x, int n) { double ans = 0; for (int i = 0; i <= n; i++) ans += f[i] * pow(x, i); return ans; } double newton(double *f, int n) { double fd[N]; for (int i = 0; i < n; i++) fd[i] = f[i + 1] * (i + 1); double x = -25.0; for (int i = 0; i < 100; i++) x = x - cal(f, x, n) / cal(fd, x, n - 1); return x; } void tra(double *f, double x, int n) { f[n + 1] = 0; for (int i = n; i > 0; i--) f[i] = f[i + 1] * x + f[i]; for (int i = 0; i < n; i++) f[i] = f[i + 1]; } void solve() { for (int i = 0; i < n; i++) { double x = newton(a, n - i); printf(" %.4lf", x); tra(a, x, n - i); } } int main() { int cas = 0; while (~scanf("%d", &n) && n) { for (int i = n; i >= 0; i--) scanf("%lf", &a[i]); printf("Equation %d:", ++cas); solve(); printf("\n"); } return 0; }
UVA 10428 - The Roots(牛顿迭代法),布布扣,bubuko.com
原文:http://blog.csdn.net/accelerator_/article/details/38116117