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

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7493    Accepted Submission(s): 3484


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
 

 

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　　二分搜索。
　　先要判断方程的单调性。可用 cal(0)<=y && y<=cal(100) 判断。
　　二分法不断提高精度，最后到某个精度位置输出就是正确答案。
　　注意题目给出的第一组测试数据是错的，AC代码出来的结果是1.6151，而不是1.6152。
　　代码：
 1 #include <iostream>
 2 #include <cmath>
 3 #include <iomanip>
 4 using namespace std;
 5 
 6 double cal(double x)
 7 {
 8     return 8*pow(x,4) + 7*pow(x,3) + 2*pow(x,2) + 3*x + 6;
 9 }
10 double GetAns(double y)
11 {
12     double a=0,b=100;
13     while(b-a>1e-6){
14         double mid=(a+b)/2;
15         cal(mid)<y?a=mid:b=mid;
16     }
17     return (a+b)/2;
18 }
19 int main()
20 {
21     int T;
22     cin>>T;
23     while(T--){
24         double y;
25         cin>>y;
26         if(cal(0)<=y && y<=cal(100)){
27             cout<<setiosflags(ios::fixed)<<setprecision(4);
28             cout<<GetAns(y)<<endl;
29         }
30         else
31             cout<<"No solution!"<<endl;
32     }
33     return 0;
34 }

 

Freecode : www.cnblogs.com/yym2013

posted @ 2014-04-30 12:11  Freecode#  阅读(248)  评论(0)  编辑  收藏  举报