hdu 2199 （二分）

最新推荐文章于 2025-08-11 22:14:14 发布

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最新推荐文章于 2025-08-11 22:14:14 发布

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文章标签： java php

原文链接：http://www.cnblogs.com/ccccnzb/p/3932652.html

本文介绍了一种使用二分查找法解决特定多项式方程的方法。该方程为8*x^4+7*x^3+2*x^2+3*x+6=Y，在0到100范围内寻找解。通过输入不同的Y值，程序可以返回方程的解或提示无解。

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链接：http://acm.hdu.edu.cn/showproblem.php?pid=2199

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8595 Accepted Submission(s): 3957

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

100

-4

Sample Output

1.6152

No solution!

------------------------------------------------------------------------------------------------------------、

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <stdlib.h>
 4 #include <iostream>
 5 #include <algorithm>
 6 #include <math.h>
 7 #define eps 1e-8
 8 using namespace std;
 9 
10 double fx(double x)
11 {
12     return 8*pow(x,4)+7*pow(x,3)+2*x*x+3*x+6;
13 }
14 
15 int main()
16 {
17     int n,m,i,j;
18     double y;
19     double left=0,right=100,ans;
20     scanf("%d",&n);
21     while(n--)
22     {
23         scanf("%lf",&y);
24         left=0.0,right=100.0;
25         //ans=-100000;
26         bool flag=true;
27         while(right-left>eps)
28         {
29             double mid=(left+right)/2.0;
30             if(fx(mid)-y< 0)
31                  left=mid+eps;
32             else right=mid-eps;
33             //if()flag=false;
34         }
35         if(y>fx(100) || y<fx(0))
36             printf("No solution!\n");
37         else
38             printf("%.4lf\n",left);
39     }
40 
41 }