HDU2199(二分搜索无限逼近)

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4276    Accepted Submission(s): 1989

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

 

Recommend

lcy

 

好吧这题真的有点跪了，做了那么久，用二分来无限逼近一个准确值，则将二分循环的结束条件设置为小于该精度，为了准确输出，自然要把该精度设置得更加细。

 

#include <cstdio>
#include <cmath>

double result(double m)
{
double s=8*pow(m*1.0,4)+7*pow(m*1.0,3)+2*m*m+3*m+6;
return s;
}
int main()
{
int t;double y;
scanf("%d",&t);
  while (t--)
  {
   scanf("%lf",&y);
   double mid2;
   if(y<6||y>result(100)) puts("No solution!");
   else
   {
    double min2=1.0,max2=100.0;
    while ((max2-min2)>1e-6)
    {
      mid2=(max2-min2)/2+min2;
      double sum=result(mid2);
      if (sum==y) break;
      if (sum<y) min2=mid2;
      else max2=mid2;
     }
     printf("%.4f\n",mid2);
   }
  }
return 0;
}