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()<=y && y<=cal() 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：

 #include <iostream>
 #include <cmath>
 #include <iomanip>
 using namespace std;
 
 double cal(double x)
 {
     return *pow(x,) + *pow(x,) + *pow(x,) + *x + ;
 }
 double GetAns(double y)
 {
     double a=,b=;
     while(b-a>1e-){
         double mid=(a+b)/;
         cal(mid)<y?a=mid:b=mid;
     }
     return (a+b)/;
 }
 int main()
 {
     int T;
     cin>>T;
     while(T--){
         double y;
         cin>>y;
         if(cal()<=y && y<=cal()){
             cout<<setiosflags(ios::fixed)<<setprecision();
             cout<<GetAns(y)<<endl;
         }
         else
             cout<<"No solution!"<<endl;
     }
     return ;
 }

Freecode : www.cnblogs.com/yym2013