C语言数组学习的困惑，望各位大神解答

[夜鸣]

写一个程序要求能输出重复的数字，不知道哪里出了问题，输出的结果一直是一堆零，求指点，下面是所写程序：


#include<stdio.h>
int main(void)
{
int digit_seen[10] = { 0 }, repeated_digit[10] = { 0 };
int digit,i=0,j=0;
long n;
printf("Enter a number: ");
scanf_s("%ld", &n);
while (n > 0) {
digit = n % 10;
digit_seen[digit] = digit;
if (digit_seen[digit] == digit)
{
repeated_digit[j] = digit;
j++ ;
}
n /=10;
}

if (j > 0)
{
printf("Repeated digit\n");
for (i = 0; i <= j; i++)
{
printf(" %d", repeated_digit[j]);
}
}
else
printf("No repeated digit\n");
return 0;
}

[636f6c696e]

1. 你的算法不对，if循环永远成立 2. 最后输出的时候应该是printf(" %d", repeated_digit[i]);

[自信男孩]

#include<stdio.h>
int main(void)
{
    int digit_seen[10] = { 0 };//, repeated_digit[10] = { 0 };
    int digit,i=0, flag = 0;//j=0;
    long n;
    printf("Enter a number: ");
    //scanf_s("%ld", &n);
    scanf("%ld", &n);
    while (n > 0) {
        digit_seen[n % 10]++;
        n /= 10;
        /*
        digit = n % 10;
        digit_seen[digit] = digit;
        if (digit_seen[digit] == digit)
        {
            repeated_digit[j] = digit;
            j++ ;
        }
        n /=10;
        */
    }

    for (i = 0; i < 10; i++)
        if (digit_seen[i] > 1) {
            if (flag == 0)
                printf("Repeated digit\n");
            printf("%d ", i);
            flag = 1;
        }
    if (flag == 0)
        printf("No repeated digit\n");
    else
        putchar(10);

/*
    if (j > 0)
    {
        printf("Repeated digit\n");
        for (i = 0; i <= j; i++)
        {
            printf(" %d", repeated_digit[j]);
        }
    }
    else
        printf("No repeated digit\n");
*/
    return 0;
}

供参考
一个数组就可以完成了！