HDU 2199 Can you solve this equation? 二分

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15347    Accepted Submission(s): 6829

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!

 

Author

Redow

 

Recommend

lcy

 

题目大意：输入一个数T，接下来输入T组数据(Y)，判断是否存在X(0<=X<=100),使得方程8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y成立？？

 

思路：二分妥妥的。

注意解出X超过100和小于0不存在，还有精度问题。。。

 

#include <iostream>
#include <math.h>
using namespace std;
double v(double x)
{
    return (8*pow(x,4)+7*pow(x,3)+2*pow(x,2)+3*x+6);
}
double f(double x,double z, double y)
{
double mid;
//int t=50;
    while((y-z)>1e-10)//用条件不太好用哈，可以用循环次数t--
    {
        mid=(z+y)/2;
        if(v(mid)<x)
            z=mid+1e-10;
        else
            y=mid-1e-10;
    }
    return mid;
    
}
int main()
{
    int n;
    double a;
    scanf("%d", &n);
    while(n--)
    {
        scanf("%lf", &a);
        if(a<6||a>807020306||fabs(f(a,0,100))<1e-4)
        printf("No solution!
");
        else
        printf("%.4lf
", f(a,0,100));
    }
}