hdu 2199 Can you solve this equation?(二分)

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10943    Accepted Submission(s): 5046

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

2100-4

 

Sample Output

1.6152No solution!

 

Author

Redow

题目分析：精度比较恶心的二分，只要不超时，精度尽量开小就对了，涨姿势了
 

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <cmath>#define eps 1e-12#define EPS 1e-9using namespace std;double f ( double x ){    return 8*x*x*x*x + 7*x*x*x + 2*x*x + 3*x + 6.0;}bool check ( double x , double n ){    if ( f(x)+eps < n ) return true;}bool equal ( double a , double b ){    if ( fabs ( a -b ) < eps ) return true;    else return false;}double search ( double n ){    double left = 0.0 , right = 100.0 , mid;    while ( !equal ( left , right ))    {        mid = (left + right) / 2.0;        if ( check ( mid , n ) ) left = mid;        else right = mid;    }    return mid;}int t;double n;int main ( ){    scanf ( "%d" , &t );    while ( t-- )    {       scanf ( "%lf" , &n );       if ( n < f(0) || n > f(100 ) )       {           puts ( "No solution!" );           continue;       }       double ans = search ( n );       printf ("%.4lf\n" , ans );    }}