c - 该算法存在一些缺陷，无法通过名为 "partly calculate A add B"的所有OJ测试

Question

OJ 的问题是:

Question: Give three positive integer like A Da B Db and calculate Pa+Pb. And the rule likes Example: A = 3862767, Da = 6,so Pa = 66, because there are two number "6" in integer A.

Input Format: Enter A Da B Db in one line, and use space to split them, and 0<A,B<10^10

Output Format: Show the value of Pa+Pb in one line

Input Example 1: 3862767 6 13530293 3

Output Example 1: 399

Input Example 2: 3862767 1 13530293 8

Output Example 2: 0

我的问题:我已经通过了五项测试中的四项，但不知道为什么我的代码无法全部通过并且编译器是 gcc 4.7.2

我的代码是:

int fun(int a, int b) {
    int n = 0;
    short i = 1,k;
    do {
        k = a % 10;
        if (k == b) {
            n += i*k;
            i *= 10;
        }
    } while ((a /= 10) != 0);
    return n;
}

int main() {
    int a1, a2, a3, a4;
    scanf("%d %d %d %d", &a1, &a2, &a3, &a4);
    printf("%d", fun(a1, a2) + fun(a3, a4));
    return 0;
}

Answer 1

是的，我不应该使用 short 和 int，更改为 long long 后测试已通过。@david-eisenstat