HDU2199 Can you solve this equation?

最新推荐文章于 2021-04-09 16:43:57 发布

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本文提供了解决方程8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 = Y的算法，并通过输入实数Y来确定解是否存在于0到100之间，若存在则输出精确到小数点后四位的解，否则输出无解！。

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5667 Accepted Submission(s): 2681

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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代码如下：

#include<stdio.h>
#include<math.h>
double f(double x)
{
    return 8*pow(x,4.0)+7*pow(x,3.0)+2*pow(x,2.0)+3*x+6;
}
int main()
{
   double l,h,m,ans,y;
   int n;
   while(scanf("%d",&n)!=EOF)
   {
     while(n--)
     {
         scanf("%lf",&y);
           if(f(0)<y&&y<=f(100))
          {
              l=0;h=100;
              while(h-l>1e-6)
              {
                  m=(l+h)/2;
                  ans=f(m);
                  if(ans>y)
                  {
                     h=m-1e-7;
                  }
                  else
                  {
                      l=m+1e-7;
                  }
              }
              printf("%.4lf\n",(l+h)/2);
          }
           else
           printf("No solution!\n");
     }
   }
   return 0;
}