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