解题思路：给出一个方程 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y，求方程的解。 首先判断方程是否有解，因为该函数在实数范围内是连续的，所以只需使y的值满足f(0)<=y<=f(100),就一定能找到该方程的解，否则就无解。 然后是求解过程， 假设一个区间[a,b],mid=(a+b)/2,如果f(a)*f(b)<0,那么函数f(x)在区间[a,b]至少存在一个零点，如果f(a)<0,说明0点在其右侧，那么将a的值更新为当前mid的值，如果f(a)>0,说明0点在其左侧，将b的值更新为mid的值。画出图像更好分析。

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)  

  Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 9490    Accepted Submission(s): 4382

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!

#include
     
      #include
      
       double f(double x){	return 8*x*x*x*x+7*x*x*x+2*x*x+3*x+6;}int main(){	int ncase;	double y,ans,left,right,mid;	scanf("%d",&ncase);	while(ncase--)	{		scanf("%lf",&y);		ans=0;		left=0;	    right=100;	    if(f(0)<=y&&y<=f(100))	    {		while(right-left>0.000000001)		{					{				mid=(left+right)*0.5;				ans=f(mid);				if(ans-y<0)				left=mid;				else				right=mid;				}								}		printf("%.4lf\n",mid);	}		else		printf("No solution!\n");	}}