Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem J
The Roots
Input: standard input
Output: standard output
Time Limit: 15 seconds

A polynomial equation has the following form:

                 a_nxⁿ + a_n-1x^n-1 + a_n-2x^n-2 + … + a₁x¹+ a₀ = 0

Here X is variable and a_n, a_{n-1 …} etc are known as coefficient.  So to specify a polynomial equation we need the values of the coefficients only. The roots of a polynomial equation are the values of X for which the value of LHS is zero. In this problem you will have to find the roots of a polynomial equation. You can assume that a polynomial equation of degree n should have n real roots and all the roots are strictly different.

Input

The input file contains less than 5001 lines of input.

Each line contains an integer N (N<=5) followed by (N+1) floating point numbers. Here N is the order of the polynomial equation. The next (N+1) numbers denote the values of a_n, a_n-1, a_n-2  …, a₁ and a₀. These absolute values of these coefficients will be less than 1e9 or 1000000000 and the absolute values of the roots will be less than 25.  So each line contains information about one polynomial equation.

Input is terminated by a line where N=0.

Output

For each line of input produce one line of output. This line starts with the string “Equation S:” here S is the output serial number as shown in the sample output. This string is followed by N floating point numbers, which are the roots of the corresponding input equation. The roots should be printed in ascending order of their values and rounded up to four digits after the decimal point. All the root values are preceded by a single space. The judge input is designed in such a way that if your precision is relatively low (values less than 1e-10 are considered as zero) you will face no precision errors. Of course your solution must be correct and you must be well aware of the limitations of different root finding methods.

Sample Input
2 1 -9.5750000000 -179.5585140000

4 1 53.3120000000 958.2510390000 6677.1763593480 15733.6254955064

0

Sample Output

Equation 1: -9.4420 19.0170

Equation 2: -22.8060 -18.1170 -6.7350 -5.6540

Problem setter: Shahriar Manzoor, ACM Valladolid Online Judge Team

牛頓法 x' = x - f(x)/f'(x) 然後去逼近解。

思考會不會有 -0.0000 的輸出要修正, 導致精準度上要做調整。

找到一個解之後, 用多項式的直式除法(而非長除法)去降次。

#include <stdio.h>
#include <algorithm>
#include <math.h>
using namespace std;
int main() {
    int n, i, j, k;
    int cases = 0;
    while(scanf("%d", &n) == 1 && n) {
        double a[10] = {};
        for(i = n; i >= 0; i--)
            scanf("%lf", &a[i]);
        double ret[10];
        int m = n;
        for(i = 0; i < m; i++) {
            double b[10] = {}; // f'(x)
            for(j = 0; j <= n; j++)
                b[j] = a[j+1]*(j+1);
            double x = 25, tx;//max_value
            if(i)   x = ret[i-1];
            while(true) {
                double fx = 0, ffx = 0;
                for(j = 0; j <= n; j++) {
                    fx += a[j]*pow(x, j);
                    ffx += b[j]*pow(x, j);
                }
                tx = x - fx/ffx;
                if(fabs(fx) < 1e-8)
                    break;
                x = tx;
            }
            ret[i] = x;
            for(j = n; j >= 0; j--)
                a[j] = a[j] + a[j+1]*x;
            for(j = 0; j <= n; j++)
                a[j] = a[j+1];
            n--;
        }
        printf("Equation %d:", ++cases);
        n = m;
        sort(ret, ret+n);
        for(i = 0; i < n; i++)
            printf(" %.4lf", ret[i]);
        puts("");
    }
    return 0;
}
/*
2 1 -9.5750000000 -179.5585140000

4 1 53.3120000000 958.2510390000 6677.1763593480 15733.6254955064

0
*/