Equation (分类讨论)

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原创最新推荐文章于 2026-06-24 11:05:46 发布 559 阅读 AI 写同款· GEO 优化 ›

机器学习（四）：w·x+b模型（1）假设输入空间x∈Rnx\in R^n，对于分类问题，我们使用的假设空间为H={h=sign(wTx+b)|w∈Rn,b∈R}H=\{h=sign(w^Tx+b)| w\in R^n, b\in R\} 对于回归问题，我们使用的假设空间为H={h=wTx+b|w∈Rn,b∈R}H=\{h=w^Tx+b|w\in R^n, b\in R\} 其实际意义也比较简单：对于分类问题，用一个超平面将特征阅读详情

Equation

You are given an equation:

Ax² + Bx + C = 0.

Your task is to find the number of distinct roots of the equation and print all of them in ascending order.

Input

The first line contains three integer numbers A, B and C ( - 10⁵ ≤ A, B, C ≤ 10⁵). Any coefficient may be equal to 0.

Output

In case of infinite root count print the only integer -1. In case of no roots print the only integer 0. In other cases print the number of root on the first line and the roots on the following lines in the ascending order. Print roots with at least 5 digits after the decimal point.

Example

Input

1 -5 6

Output

2
2.0000000000
3.0000000000

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
int main(){
    double a,b,c;
    cin >> a >> b >> c;
    if(a == 0 && b == 0){
        if(c == 0) printf("-1\n");
        else printf("0\n");
    }
    else if(a == 0 && c == 0){
        printf("1\n0\n");
    }
    else if(b == 0 && c == 0){
        printf("1\n0\n");
    }
    else if(a == 0){
        printf("1\n");
        printf("%.10f\n",-c/b);
    }
    else{
        double diat = b*b-4*a*c;
        if(diat < 0) printf("0\n");
        else if(diat == 0){
            printf("1\n");
            printf("%.10f\n",(-b/(2*a)));
        }
        else{
            printf("2\n");
            double x1 = (-b-sqrt(diat))/(2*a);
            double x2 = (-b+sqrt(diat))/(2*a);
            if(x1 > x2) swap(x1,x2);
            printf("%.10f\n%.10f\n",x1,x2);
        }
    }
    return 0;
}

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