UVA 10428 - The Roots

题目链接

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;
}