hdu 2199 Can you solve this equation?(二分搜索)

Can you solve this equation?

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!
 
：：很容易知道y=8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 单调递增，那么我们就可以对[0,100]进行二分搜索，找出满足条件的数
 
代码：
   1: #include <iostream>
   2: #include <cstdio>
   3: #include <cstring>
   4: #include <algorithm>
   5: #include <cmath>
   6: using namespace std;
   7:  
   8: double f(double x){
   9:     return 8*x*x*x*x+7*x*x*x+2*x*x+3*x+6;
  10: }
  11:  
  12: int main()
  13: {
  14:     int T,y;
  15:     scanf("%d",&T);
  16:     while(T--)
  17:     {
  18:         scanf("%d",&y);
  19:         if(f(0)>y||f(100)<y)//f(0)>y或f(100)<y则无解
  20:         {
  21:             printf("No solution!\n");
  22:             continue;
  23:         }
  24:         double l=1.0,r=100.0;
  25:         while(r-l>1e-6)
  26:         {
  27:             double m=(l+r)/2;
  28:             if(f(m)>y) r=m-1e-7;
  29:             else l=m+1e-7;
  30:         }
  31:         printf("%.4lf\n",(l+r)/2);
  32:     }
  33:     return 0;
  34: }

hdu 2199 Can you solve this equation?(二分搜索),布布扣,bubuko.com

hdu 2199 Can you solve this equation?(二分搜索)

原文：http://www.cnblogs.com/zyx1314/p/3608957.html