hdu 2199 Can you solve this equation? 二分

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6763 Accepted Submission(s): 3154

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!

Author

Redow

Recommend

lcy | We have carefully selected several similar problems for you: 2899 2289 2298 2141 3400

 1 /*
 2 对于精度，我表示囧。
 3 我以为，保留4位小数，就到1e-5就可以了。
 4 
 5 */
 6 
 7 #include<iostream>
 8 #include<stdio.h>
 9 #include<cstring>
10 #include<cstdlib>
11 #include<math.h>
12 using namespace std;
13 
14 double fun(double x)
15 {
16     return 8*x*x*x*x+7*x*x*x+2*x*x+3*x+6;
17 }
18 void EF(double l,double r,double Y)
19 {
20     double mid;
21     while(r-l>1e-7)
22     {
23         mid=(l+r)/2;
24         double ans=fun(mid);
25         if( ans >Y )
26             r=mid-1e-8;
27         else l=mid+1e-8;
28     }
29     printf("%.4lf
",(l+r)/2);
30 }
31 int main()
32 {
33     int T;
34     double Y;
35     scanf("%d",&T);
36     {
37         while(T--)
38         {
39             scanf("%lf",&Y);
40             if( fun(0.0)>Y || fun(100.0)<Y)
41                 printf("No solution!
");
42             else
43                  EF(0.0,100.0,Y);
44         }
45     }
46     return 0;
47 }