HDU 2199

最新推荐文章于 2021-01-14 20:42:38 发布

原创最新推荐文章于 2021-01-14 20:42:38 发布 · 246 阅读

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Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 20197 Accepted Submission(s): 8892

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

Sample Output

  
   1.6152
No solution!
  
  

  
  

  
  
   题意：利用二分搜索求方程的解，只需输出一个实数（精确到小数点后4位），这是方程的解，或者“没有解”！如果x在0和100之间，方程没有解的话。
  
  
   注：原型：extern float pow(float x, float y);
  
    用法：#include <math.h>
  
    功能：计算x的y次幂。
  
    说明：x应大于零，返回幂指数的结果。
   #include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
const double esp=1e-10;//十的负十次方
double f(double x)
{
    return 8.0*pow(x,4)+7*pow(x,3)+2*pow(x,2)+3*x+6;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        double y;
        scanf("%lf",&y);
        double left=0.0;
        double right=100.0;
        if(f(left)>y||f(right)<y)
        {
            printf("No solution!\n");
            continue;
        }
        while(right-left>esp)
        {
            double mid=(left+right)/2;
            if(f(mid)>y)
            {
                right=mid;
            }
            else
            {
                left=mid;
            }

        }
        printf("%.4lf\n",right);

    }
    return 0;
}