HDOJ 2199 Can you solve this equation?(二分查找)

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12794    Accepted Submission(s): 5715

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!
   
   
   
   

 

题目说x的值在[0,100]内，所以判断得当y<6 || y>807020306时，输出No solution!   然后进行二分查找

 

代码如下：

 

#include<cstdio>
#include<cmath>
int find(double x,double y)
{
	if(y-(8.0*pow(x,4)+7.0*pow(x,3)+2.0*pow(x,2)+3.0*x+6)<0)
	   return 1;
	else
	   return 0;
}

int main()
{
	int t;
	double y,mid,L,R;
	scanf("%d",&t);
	while(t--)
	{
		L=0;
		R=100;
		scanf("%lf",&y);
		if(y<6||y>807020306)
		   printf("No solution!\n");
		else
		{
			while(R-L>1e-8)
			{
				mid=(R+L)/2.0;
				if(find(mid,y))//返回值为1的时候表示x的值取大了，则继续在小于mid的区间寻找 
				   R=mid;
				else
				   L=mid;
			}
			printf("%.4lf\n",mid);
		}
	}
	return 0;
}