HDU 2199 (Binary)

Link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 2199

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 8595 accepted submission (s): 3957

Problem descriptionnow, 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.

 

Inputthe 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 );

 

Outputfor each test case, You shocould 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 Input2100-4

 

Sample output1.6152no solution! Average ------------------------------------------------------------------------------------------------------------,

 1 #include <stdio.h> 2 #include <string.h> 3 #include <stdlib.h> 4 #include <iostream> 5 #include <algorithm> 6 #include <math.h> 7 #define eps 1e-8 8 using namespace std; 9 10 double fx(double x)11 {12     return 8*pow(x,4)+7*pow(x,3)+2*x*x+3*x+6;13 }14 15 int main()16 {17     int n,m,i,j;18     double y;19     double left=0,right=100,ans;20     scanf("%d",&n);21     while(n--)22     {23         scanf("%lf",&y);24         left=0.0,right=100.0;25         //ans=-100000;26         bool flag=true;27         while(right-left>eps)28         {29             double mid=(left+right)/2.0;30             if(fx(mid)-y< 0)31                  left=mid+eps;32             else right=mid-eps;33             //if()flag=false;34         }35         if(y>fx(100) || y<fx(0))36             printf("No solution!\n");37         else38             printf("%.4lf\n",left);39     }40 41 }

View code

 

HDU 2199 (Binary)