HDU Problem 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): 16602    Accepted Submission(s): 7371

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!

 

题中说解在0~100之间，所以当输入的值小于x=0时的值大于x=100时候的值是不存在的，其他情况都有解，所以，对于其他情况，直接进行二分。

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

const double sp = 1e-5;
double ans(double x) {
    return 8*pow(x, 4.0) + 7*pow(x, 3.0) + 2*pow(x, 2.0) + 3*x + 6;
}

double lower_bound(double x) {
    double l = 0, r = 100;
    double mid = (l + r)/2;
    while (abs(ans(mid) - x) > sp) {
        mid = (l + r)/2;
        if (ans(mid) > x) {
            r = mid;
        }
        else {
            l = mid;
        }
    }
    return mid;
}
int main() {
    int t;
    double n;
    scanf("%d", &t);
    while (t--) {
        scanf("%lf", &n);
        if (n < 6 || n > ans(100.0))
            printf("No solution!
");
        else printf("%.4lf
", lower_bound(n));
    }
    return 0;
}