蓝桥杯算法训练VIP-一元三次方程求解，简单问题复杂化解法！！！

解题思路:

注意事项:

参考代码:

#include"bits/stdc++.h"
using namespace std;
int main(){
    // 定义变量a, b, c, d用于存储输入的系数
    double a,b,c,d;
    // 初始化计数器g为0，用于记录找到的根的数量
    int g=0;
    // 定义变量o, p, s1, s2, s3, u, y, t用于后续计算和存储结果
    double o,p,s1,s2,s3,u,y,t;
    // 从标准输入读取四个系数a, b, c, d
    cin>>a>>b>>c>>d;
    // 遍历x1从-100到100的所有整数
    for(int x1=-100;x1<=100;x1+=1){
        // 遍历y1从-100到100的所有整数
        for(int y1=-100;y1<=100;y1+=1){
            // 确保x1小于y1
            if(x1<y1){
                // 检查函数在x1和y1处的值是否异号
                if((a*x1*x1*x1+b*x1*x1+c*x1+d)*(a*y1*y1*y1+b*y1*y1+c*y1+d)<0){
                    // 初始化l和k为当前的x1和y1
                    double l=x1,k=y1;
                    // 在区间[l, k]内以步长0.01进行迭代
                    for(double x2=l;x2<k;x2+=0.01){
                        // 检查函数值是否接近于零（误差小于0.0001）
                        if(fabs(a*x2*x2*x2+b*x2*x2+c*x2+d)<0.0001){
                            // 如果这是第一个找到的根
                           if(g==0){
                               // 将该根赋值给s1
                               s1=x2;
                               // 增加计数器g
                               g++;
                               // 更新o和p为当前根的值
                               p=x2;
                               o=x2;
                               // 跳出当前循环
                               break;
                           }
                           // 如果这是第二个找到的根且与第一个根的距离大于等于1.00
                           else if(g==1&&fabs(x2-o)>=1.00){
                               // 将该根赋值给s2
                               s2=x2;
                               // 增加计数器g
                               g++;
                               // 更新o为当前根的值
                               o=x2;
                               // 跳出当前循环
                               break;
                           }
                           // 如果这是第三个找到的根且与前两个根的距离都大于等于1.00
                           else if(g==2&&fabs(x2-o)>=1.00&&fabs(x2-p)>=1.00){
                               // 将该根赋值给s3
                               s3=x2;
                               // 增加计数器g
                               g++;
                               // 更新o为当前根的值
                               o=x2;
                               // 跳出当前循环
                               break;
                           }
                           // 如果已经找到了三个根，跳转到标签aaa
                           else if(g==3){
                               goto aaa;
                           }
                        }
                    }
                }
            }
        }
    }
    // 标签aaa，用于结束嵌套循环
    aaa:
    // 找出s1, s2, s3中的最大值并赋值给y
    y=max(max(s1,s2),s3);
    // 找出s1, s2, s3中的最小值并赋值给u
    u=min(min(s1,s2),s3);
    // 根据最大值和最小值确定中间值t
    if(y==s1&&u==s3||y==s3&&u==s1){
        t=s2;
    }
    else if(y==s2&&u==s1||y==s1&&u==s2){
        t=s3;
    }
    else{
        t=s1;
    }
    // 输出结果，保留两位小数
    cout<<fixed<<setprecision(2)<<u<<" "<<t<<" "<<y;
    return 0;
}