P1024 [NOIP2001 提高组] 一元三次方程求解

题目链接

https://www.luogu.com.cn/problem/P1024

题目思路

二分 ：根据题意，两根乘积 < 0，两根之间必有根，并且两根之差 >= 1，所以以1为大区间，若两根乘积为负则二分寻找

暴力 ：因为保留两位小数，则每次缩小区间0.01，直接寻找乘积为0的两根，其平均数的近似值即为根

题目代码

二分

#include <iostream>
#include <algorithm>

using namespace std;
double a, b, c, d;

double f(double x)
{
    return a * x * x * x + b * x * x + c * x + d;
}

int main()
{
    double x1, x2;
    double l, r;
    scanf("%lf%lf%lf%lf", &a, &b, &c, &d);
    for(double i = -100; i < 100; i ++)
    {
        l = i, r = i + 1;
        x1 = f(l), x2 = f(2);
        if(x1 == 0) printf("%.2f ", l);
        if(x1 * x2 < 0)
        {
            while(r - l >= 1e-3)
            {
                double mid = (l + r) / 2;
                if(f(mid) * f(r) <= 0) l = mid;
                else r = mid;
            }
            printf("%.2f ", r);
        }
    }
    return 0;
}

暴力

#include <iostream>
#include <algorithm>

using namespace std;


int main()
{
    double a, b, c, d;
    scanf("%lf%lf%lf%lf", &a, &b, &c, &d);
    for(double i = -100; i <= 100; i += 0.001)
    {
        double j = i + 0.001;
        double ans1 = a * i * i * i + b * i * i + c * i + d;
        double ans2 = a * j * j * j + b * j * j + c * j + d;
        if(ans1 >= 0 && ans2 <= 0 || ans1 <= 0 && ans2 >= 0)
        {
            double x = (i + j) / 2;
            printf("%.2f ", x);
        }
    }
    return 0;
}