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 (*a*a*a*a+*a*a*a+*a*a+*a+);
 }
 int main(void)
 {
     int t;
     double n;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%lf",&n);
         if(n< || n>fac()){
             puts("No solution!");continue;
         }
         double l=;
         double r=;
         while(r-l>e){
             double mid=(l+r)/;
             if(fac(mid)>n) r=mid;
             else l=mid;
         }
         printf("%.4lf\n",r);
     }
     return ;
 }

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