hdu 2199 二分法求单调函数方程的解

首先，判断f(x)是单调递增的。 然后用， double 型的 二分模板

 

题目来源：  

http://acm.hdu.edu.cn/showproblem.php?pid=2199

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7258    Accepted Submission(s): 3362

Problem Description

Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 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!

 

 

代码如下：

#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <stack>
#include <iostream>
#include <stdio.h>
#include <string>
#include <string.h>
#include <algorithm>
#include <stdlib.h>
#include <vector>
#include <set>
#include <math.h>
#include <cmath>
#include <map>
#include <stack>
#include <queue>
using namespace std ;

typedef long long LL ;
const double EPS=1e-8;

double f(double x)
{
    return 8*pow(x,4)+7*pow(x,3)+2*pow(x,2)+3*x+6;
}
double Bi_Search(double y)
{
    double left,right,mid;
    left=0,right=100;
    while(right-left>EPS)
    {
    mid=(left+right)*0.5;
    if(f(mid)<y)
        left=mid;
    else right=mid ;
    }
    return (left+right)*0.5;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
    double y;
    cin>>y;
    if(y>=f(0)&&y<=f(100))
    printf("%.4lf\n",Bi_Search(y));
    else cout<<"No solution!"<<endl;
    }
return 0;
}