The sum of 10 natural numbers within k that are divisible by 13 or 17

code show as below:

#include<stdio.h>
int fun(int k){
    
    
	int m=0,mc=0;
	int count = 0;
	printf("\nk以内的10个能被13或17整除的自然数:");
	while((k >= 2)&&(mc < 10)){
    
    
		if((k % 13 == 0)||(k % 17 == 0)){
    
    
			m = m+ k;
			mc ++;
			count ++;
			//输出能被13或17整除的自然数
			printf("\n\n");
			printf("\tk = %d",k);
		}
		k --;
	}
	//500以内的10个能被13或17整除的自然数的和
	printf("\n\nk以内的10个能被13或17整除的自然数的和： m = %d\n\n",m);
	return m;
}
int main(){
    
    
	fun(500);
}