题目链接：uva 10428 - The Roots

题目大意：给定一个n次一元多项式，求出所有解。

解题思路：牛顿迭代法，对于任意给定x，通过牛顿迭代法可以趋近距离x最近的解x0。每次找到一个解后，用多项式除法除掉x?x0后继续求解。

牛顿迭代法：xi+1=xi?f(x)f′(x)

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 10;

int N;
double a[maxn];

void div (double* f, double x, int n) {
    f[n+1] = 0;
    for (int i = n; i >= 0; i--)
        f[i] += f[i+1] * x;

    for (int i = 0; i < n; i++)
        f[i] = f[i+1];
}

double func (double* f, double x, int n) {
    double ret = 0, u = 1;
    for (int i = 0; i <= n; i++) {
        ret += f[i] * u;
        u *= x;
    }
    return ret;
}

double newton (double* f, int n) {
    double fd[maxn];
    for (int i = 0; i < n; i++)
        fd[i] = f[i+1] * (i+1);
    double x = -100;
    for (int i = 0; i < 100; i++)
        x -= func(f, x, n) / func(fd, x, n-1);
    return x;
}

void solve () {
    for (int i = 0; i < N; i++) {
        double x = newton(a, N-i);
        printf(" %.4lf", x);
        div(a, x, N-i);
    }
}

int main () {
    int cas = 1;
    while (scanf("%d", &N) == 1 && N) {
        for (int i = N; i >= 0; i--)
            scanf("%lf", &a[i]);
        printf("Equation %d:", cas++);
        solve();
        printf("\n");
    }
    return 0;
}