HDOJ--2199Can you solve this equation?!!!！二分法

原题链接：

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!

题目大意：题目大意就是给你一个函数

Now, here is a fuction:
  F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100) ，然后给你F（X）的数值。计算在0-100区间内的解，如果解不在0-100区间内，那么输出No solution!

思路：二分，第一次尝试二分法，查了很多人的资料，初步理解了二分法。

代码：

#include<stdio.h>
#include<math.h>
double f(double x)
{
	return 8*pow(x,4.0)+7*pow(x,3.0)+2*pow(x,2.0)+3*x+6;
}
//函数求导可以发现，在0-100是单调递增的 
int main()
{
	double max=f(100);      //最大值 
	double min=f(0);		//最小值 
	int T;
	double Y,result,l,r;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%lf",&Y);
		if(Y>max||Y<min)	//不在区间的判定 
		{
			printf("No solution!\n");
			continue;
		}
		else
		{
			l=0.0;r=100.0;
			while(r-l>1e-10)
			{
				result=f((l+r)/2);		//二分法 
				if(result<Y)
				{
					l=((l+r)/2);
				}
				else	r=((l+r)/2);
			}
			printf("%.4lf\n",(l+r)/2);
		}
	}
	return 0;
}