【二分解方程】hdu 2199 Can you solve this equation?

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20362    Accepted Submission(s): 8939

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!

 
///AC代码

#include <iostream>#include <set>#include <map>#include <stack>#include <cmath>#include <queue>#include <cstdio>#include <bitset>#include <string>#include <vector>#include <iomanip>#include <cstring>#include <algorithm>#include <functional>#define PI acos(-1)#define eps 1e-8#define inf 0x3f3f3f3f#define debug(x) cout<<"---"<<x<<"---"<<endltypedef long long ll;using namespace std;double y;double f(double x){    return 8 * x * x * x * x + 7 * x * x * x + 2 * x * x + 3 * x + 6 - y;}int main(){    int t;    cin >> t;    while (t--)    {        double left = 0;        double right = 100;        double mid = 50;        cin >> y;        if (f(0) < 0 && f(100) < 0)        {            cout << "No solution!" << endl;        }        else if (f(0) > 0 && f(100) > 0)        {            cout << "No solution!" << endl;        }        else        {            while (right - left > 1e-8)            {                mid = (left + right) / 2;                if (f(mid) > 0)                {                    right = mid;                }                else if (f(mid) < 0)                {                    left = mid;                }            }            printf("%0.4f\n", left);        }    }    return 0;}