HDU-2199（二分）

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14802    Accepted Submission(s): 6597

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!

 

题意：给一个y值，求方程在0到100之间有没有解。

思路：二分。

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

#define eps 0.000001

int t;
double y;

double f(double x){
    return 8.0*x*x*x*x+7.0*x*x*x+2.0*x*x+3.0*x+6.0;
}

//递归版eps二分
double b_srch(double begin, double end){
    double mid = (begin + end) / 2;
    double ret;
    double result = f(mid);
    if(fabs(result - y) < eps){
        return mid;
    }
    else if(result > y){
        ret = b_srch(begin, mid);
    }
    else if(result < y){
        ret = b_srch(mid, end);
    }
    return ret;
}

//循环版eps二分
double b_srch2(double begin, double end){
    double mid = (begin + end) / 2;
    while(fabs(f(mid) - y) > eps){
        if(f(mid) > y){
            end = mid;
        }
        else{
            begin = mid;
        }
        mid = (begin + end) / 2;
    }
    return mid;
}

int main()
{
    scanf("%d", &t);
    while(t--){
        scanf("%lf", &y);
        if(y < 6 || y > 807020306){
            printf("No solution!\n");
        }
        else{
            double result = b_srch(0, 100);
            printf("%.4f\n", result);
        }
    }
    return 0;
}