问题 : 方程求零点

素颜马尾好姑娘i 2022-06-10 05:29 182阅读 0赞

题目描述

设函数f(x)是在区间\[a,b\]内的连续函数，且f(a)f(b)<0, 根据Role定理，f(x)在区间内必存在零点。已知f(x)=x3-x-1, 输入区间端点的值，令精确度为eps=10-5, 判断是否存在零点，如果存在，输出该零点的近似值，否则输出No zero point.  
输入  
输入包括若干行，表示该区间端点  
输出  
每行对应每个区间的计算结果，如果根存在，保留5位小数。对于有根区间，如果|f(x\*)|<=eps，则x\*为零点的近似值。  
样例输入  
\-1 0  
0 1  
1 1.5  
样例输出  
No zero point in area(-1.00000,0.00000)  
No zero point in area(0.00000,1.00000)  
The Zero Point is 1.32472 in area(1.00000,1.50000)

AC代码

#include <iostream>
    #include<cstdio>
    #include<algorithm>
    using namespace std;
    const double eps=0.00001;
    double f(double x){
        return (x*x*x-x-1)*1.0;
    }
    int main()
    {
        double left,right;
        while(scanf("%lf%lf",&left,&right)!=EOF){
            if(f(left)*f(right)>0){
                cout<<"No zero point in area";
                printf("(%.5f,%.5f)\n",left,right);
            }else if(f(left)==0){
                printf("The Zero Point is %.5f in area(%.5f,%.5f)\n",left,left,right);
            }else if(f(right)==0){
                printf("The Zero Point is %.5f in area(%.5f,%.5f)\n",right,left,right);
            }else{
                double _mind=(left+right)/2.0;
                double l=left,r=right;
                while((l-r)>eps||(l-r)<-eps){
                    if(f(_mind)*f(l)<0){
                        r=_mind;
                        _mind=(r+l)/2.0;
                    }else if(f(_mind)*f(r)){
                        l=_mind;
                        _mind=(l+r)/2.0;
                    }
                }
                printf("The Zero Point is %.5f in area(%.5f,%.5f)\n",_mind,left,right);
            }
        }
        return 0;
    }