Brush questions - Luogu -p1151 sub number integer

P1151 A whole number of children -C Language

1、 subject

2、 The problem solving process

(1) for the first time Don't pass , Super outrageous , It's because NO Of O Lower case
result ：

Code ：

// Luogu  P1151  A whole number of children 
#include <stdio.h>
#define MIN 10000// minimum value 
#define MAX 30000// Maximum 
int main() {
    
	int k;// Positive integer  0<K<1000
	int sub1,sub2,sub3;
	int i;
	int counter=0;// Counter   Solve the last line wrap problem 
	int temp[MAX];// Store results 
	scanf("%d", &k);
	for (i = MIN; i <= MAX; i++) {
    
		sub1 = i / 100;
		if ((sub1 % k) != 0) {
    
			continue;
		}
		sub2 = (i % 10000) / 10;
		if ((sub2 % k) != 0) {
    
			continue;
		}
		sub3 = i % 1000;
		if ((sub3 % k) != 0) {
    
			continue;
		}
		temp[counter] = i;
		counter++;
		/* if ((sub1 % k == 0) & (sub2 % k == 0) & (sub3 % k == 0)) { //printf("%d", i); temp[counter] = i; counter++; }*/
	}
	if (counter == 0) {
    
		printf("NO");
		return 0;
	}
	for (i = 0; i < counter; i++) {
    
		printf("%d", temp[i]);
		if (i < counter-1) {
    
			printf("\n");
		}
	}
	return 0;
}

(2) The second time
result ：

Code ：

// Luogu  P1151  A whole number of children 
#include <stdio.h>
#define MIN 10000// minimum value 
#define MAX 30000// Maximum 
int main() {
    
	int k;// Positive integer  0<K<1000
	int sub1,sub2,sub3;
	int a, b, c, d, e;
	int i;
	int counter=0;// Counter 
	scanf("%d", &k);
	for (i = MIN; i <= MAX; i++) {
    
		sub1 = i / 100 ;
		sub2 = (i % 10000) / 10 ;
		sub3 = i % 1000 ;
		if ((sub1%k == 0) & (sub2%k == 0) & (sub3%k == 0)) {
    
			counter++;
			printf("%d\n", i);
		}
	}
	if (counter == 0) {
    
		printf("No");
	}
	return 0;
}