HDU 1098 Ignatius's puzzle

Ignatius's puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8526    Accepted Submission(s): 5907

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
   
   
   
   

 

题目大意:已知f(x) = 5*x^13 + 13*x^5 + k*a*x;
现在给你给你k的值(k<10000),求a的最小整数值使得对于任意x都有f(x)的值都能整除65；
解题思路:" arbitrary integer x ,65|f(x)"主要是这句对于任意x都能整除65.由于是任意x都成立。所以可以用特殊值带入方程求解.
所以当x=1时,可得：f(1)=k*a+18.所以有18+k*a=i*65(i=1、2、3、4......).这样就可以从小到大枚举i,当找到一个
i使得a=(i*65-18)/k是一个整数时,该值就是满足上述的最小的a值.

#include<iostream>
#include<cstdio>
using namespace std;
int main (void)
{
    int k,a;
    while(~scanf("%d",&k))
    {
        a=0;
        for(int i=1;i<=10000;i++)
        {
            if((i*65-18)%k==0)
            {
                a=(i*65-18)/k;
                break;
            }
        }
        if(a==0)
        {
            printf("no\n");
        }
        else
        printf("%d\n",a);
    }
    return 0;
}