HDU1098 Ignatius's puzzle

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
解题思路：题意比较简单，就是输入一个数k,求最小的数a使得f(x)对于任意的x都能被65整除；
要让f(x)被65整除，可以把f(x)看作是13和5的倍数：现考虑f(x)是5的倍数，则有(13*x^5+k*a*x)%5==0 <=> x(13*x^4+k*a)%5==0,由于x取任意值，当x不能被5整除时，
即x与5互素，则13*x^4+k*a要能被5整除，由费马小定理，得(x^4) mod 5== 1,所以有(13+k*a)%5==0,所以(k*a)%5==2,同理(k*a)%13==8,可知如果k能被5整除或者能被
8整除时均找不到对应的a值，所以直接输出no,又因为先考虑(k*a)%5==2中：k%5==1、2、3、4，然后枚举a,只要满足(k*a)%13==8,即可输出a值。大致思路就是这样啦；
 
代码如下：
#include <cstdio>
int main(){
    int tab[5]={0,2,1,4,3};
    int k,a;
    while(~scanf("%d",&k)){
        if(k%5==0||k%13==0){
            printf("no
");
            continue;
        }
        for(a=tab[k%5];;a+=5){
            if(k*a%13==8){
                printf("%d
",a);
                break;
            }
        }
    }
    return 0;
}