Ultraviolet A 10428-the roots (Newton Iteration Method)

Link to the question: Ultraviolet A 10428-the roots

Given a n-times polynomial, all solutions are obtained.

Solution: Newton Iteration Method. For any given X, the Newton iteration method can be used to obtain the nearest x solution x0. After finding a solution, use polynomial division to remove X? Continue to solve the problem after x0.

Newton Iteration Method: 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;}