二分法解具有单调性的方程

解方程的手段有很多，如二分法、牛顿迭代法等等，本次介绍的是采用二分法，在使用二分法解方程时，有一点应该特别注意，就是那种“具有单调性”的函数才可以，否则是会有问题的

** HDU 2199**
Problem Description
Now,given the equation 8x^4 + 7x^3 + 2x^2 + 3x + 6 == Y,can you find its solution between 0 and 100;
Now please try your lucky.

Input
The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line has a real number Y (fabs(Y) <= 1e10);

Output
For each test case, you should just output one real number(accurate up to 4 decimal places),which is the solution of the equation,or “No solution!”,if there is no solution for the equation between 0 and 100.

Sample Input

2
100
-4

Sample Output

1.6152
No solution!

题目地址

#include <iostream>
#include<math.h>
#include<iomanip>
using namespace std;
double funct(double x);
double solve(double s,double e,double y);
int main()
{
    int t;
    cin>>t;
    double ant;
    double y;
    while(t--){
        cin>>y;
        ant=solve(0,100,y);
        if(ant!=-1) cout<<fixed<<setprecision(4)<<ant<<endl;
        else cout<<"No solution!"<<endl;

    }

    return 0;
}
double funct(double x){
    return 8*pow(x,4)+7*pow(x,3)+2*x*x+3*x+6;
}
double solve(double s,double e,double y){
double mid;
while(s<=e&&s>=0&&e<=100){
    mid=(s+e)/2;
    if(funct(0)>y||funct(100)<y) return -1;//由于函数单调递增，只需判断边界解就可确定目标范围内是否有解
    if(fabs(funct(mid)-y)<=0.0001) return mid;//解的误差符合要求，返回解
    if(funct(mid)<y) s=mid;
    else if(funct(mid)>y) e=mid;
}
return -1;//-1标记无解的情况
}

相似的一题：HDU 2899 strange function

Problem Description
Now, here is a fuction:
F(x) = 6 * x^7+8*x6+7x^3+5*x2-yx (0 <= x <=100)
Can you find the minimum value when x is between 0 and 100.

Input
The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line has only one real numbers Y.(0 < Y <1e10)

Output
Just the minimum value (accurate up to 4 decimal places),when x is between 0 and 100.

Sample Input

2
100
200

Sample Output

-74.4291
-178.8534

#include <iostream>
#include<math.h>
#include<iomanip>
using namespace std;
double funct(double x,double y);
double ff(double x,double y);
double solve(double s,double e,double y);
int main()
{
    int t;
    cin>>t;
    double ant;
    double y;
    while(t--){
        cin>>y;
        ant=solve(0,100,y);
      cout<<fixed<<setprecision(4)<<funct(ant,y)<<endl;

    }

    return 0;
}
double funct(double x,double y){
    return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*x*x-x*y;
}
double ff(double x,double y){
    return 42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*x-y;
}
double solve(double s,double e,double y){
double mid;
while(s<=e&&s>=0&&e<=100){
    mid=(s+e)/2;
    if(ff(0,y)>0||ff(100,y)<0) return 0;
    if(fabs(ff(mid,y))<=0.0001) return mid;
    if(ff(mid,y)<0) s=mid;
    else if(ff(mid,y)>0) e=mid;
}
}

该题的代码是我直接拿上一题的代码稍微修改了一下，思路其实都是一样的。应该说还是比较简单的一类题目