杭电acm 1098题

Problem Description

Ignatius is poor at math,he falls across a puzzle problem,so he has no choice but to appeal to Eddy. this problem describes that:f(x)=5*x^13+13*x^5+k*a*x,input a nonegative integer k(k<10000),to find the minimal nonegative integer a,make the arbitrary integer x ,65|f(x)if
no exists that a,then print "no".

Input

The input contains several test cases. Each test case consists of a nonegative integer k, More details in the Sample Input.

 

Output

The output contains a string "no",if you can't find a,or you should output a line contains the a.More details in the Sample Output.

 

Sample Input

11
100
9999

 

Sample Output

22 no 43

本题大意是要求对任意输入的K，找到一个最小的a，使得表达式都能整除65.

自己的做法是在一定范围内去找，1-10000之间找到当x=1时能够整除65的，然后在这基础上让x在1-10000之间去循环整除，如果能满足所有条件那么就找到了这个最小的a。否则就输出no。程序如下，已经AC....

 /*******************************************************
 杭电acm  1098 已经AC
 *******************************************************/

 #include <iostream>
 #include <cmath>
 #define fx 5*pow(x,13)+13*pow(x,5)+k*a*x
 using namespace std;

 int main(void)
 {
     int x=;
     int k;
     int a=;
     //long int fx;
     int mark=;
     int flag=;
     while(scanf("%d",&k)!=EOF)
     {
         mark=;
         flag=;
         //fx=5*pow(x,13)+13*pow(x,5)+k*a*x;
         for(a=;a<;a++)
         {
             if((+k*a)%==)
             {
                 for(x=;x<;x++)
                 {
                     if(fx%==)
                         continue;
                     else {break;mark=;}
                 }

             }
             if(((+k*a)%==)&&mark==)
             {flag=;break;}
         }
         if(mark&&flag)
         {
             cout<<a<<endl;
         }
         else cout<<"no"<<endl;

     }
     return ;
 }