hdu 2199 Can you solve this equation? (二分法)

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5999    Accepted Submission(s): 2828

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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//15MS    248K    537 B    C++    
/*
    
    题意：
        给出一条公式：
             8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y
        和Y的值，要求你在(0,100)的范围内求4位小数精确解
    
    二分法：
        先排除无解的情况，因为函数在(0,100)单调递增，最小值为6，
    最大值为fac(100)，先排除无解情况然后用二分法计算
         
*/
#include<stdio.h>
#define e 1e-7
double fac(double a)
{
    return (8*a*a*a*a+7*a*a*a+2*a*a+3*a+6);
}
int main(void)
{
    int t;
    double n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf",&n);
        if(n<6 || n>fac(100)){
            puts("No solution!");continue;
        }
        double l=0;
        double r=100;
        while(r-l>e){
            double mid=(l+r)/2;
            if(fac(mid)>n) r=mid;
            else l=mid;
        }
        printf("%.4lf\n",r);
    }
    return 0;
}

 

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