搜索1001

题目；<h1>Can you solve this equation? </h1><h5>Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)</h5><h5>Total Submission(s) : 213   Accepted Submission(s) : 79</h5><div class="panel_title" align="left">Problem Description</div><div class="panel_content">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;<br>Now please try your lucky.</div><div class="panel_bottom"> </div>
<div class="panel_title" align="left">Input</div><div class="panel_content">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);</div><div class="panel_bottom"> </div>
<div class="panel_title" align="left">Output</div><div class="panel_content">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.</div><div class="panel_bottom"> </div>
<div class="panel_title" align="left">Sample Input</div><div class="panel_content"><pre><div style="FONT-FAMILY: Courier New,Courier,monospace">2<br>100<br>-4<br></div>

 

Sample Output

1.6152<br>No solution!<br>

 

Author

Redow

 

代码；#include<stdio.h>
#include<math.h>
double fun(double x)
{
	return 8*pow(x,4)+7*pow(x,3)+2*pow(x,2)+3*x+6;
}
int main()
{
	int t;
	double y,mid;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lf",&y);
		if(fun(1)>y||fun(100)<y)
		{
			printf("No solution!\n");
		}
		else
		{
			double l=0,r=100;
			while(r-l>1e-9)
			{
				mid=(r+l)/2;
				if(fun(mid)>y) r=mid;
				else l=mid;
			}
			printf("%.4lf\n",mid);
		}
	}
	return 0;
}

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