Can you solve this equation?（HDU 2199）

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 33378    Accepted Submission(s): 13830

 

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!

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思路：二分裸题，求满足Y的x的最小值，如果mid代入>=y，那么从(l,mid)中找。反之在（mid，r）中找。

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代码上的总结：不要用cout输出保留精度，会有问题。用printf

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#include<iostream>
#include<algorithm>
#include<iomanip>
#include<cmath>
typedef long long LL;
using namespace std;
int  y;

double dsearch(double  l,double r)
{
	while(r-l>1e-8)
	{
		double mid=(l+r)/2;
		if(8*pow(mid,4)+7*pow(mid,3)+2*pow(mid,2)+3*mid+6>=y) r=mid;
		else l=mid;
	}
return r;	
}


int main(void)
{
	int  t;cin>>t;
	while(t--)
	{
		cin>>y;
		if(8*pow(0,4)+7*pow(0,3)+2*pow(0,2)+3*0+6>y||8*pow(100,4)+7*pow(100,3)+2*pow(100,2)+3*100+6<y)
		cout<<"No solution!"<<endl;
		else 
		{
			double p=dsearch(0,100);
	///		cout<<setprecision(5)<<p<<endl;  做题输出精度不推荐cout，这样交上去是wa的
			printf("%.4f\n",p); 
		}
	}
return 0;	
}