全网最简单的A+B Problem解法？

进制转换秒了！！！！！（全网A+B Problem最简易做法）

#include <bits/stdc++.h>
using namespace std;
int A[30], B[30], a, b;
int check(int mod) {
    int x = a, y = b, ta = 0, tb = 0;
    while (x) {
        A[++ta] = x % mod;
        x /= mod;
    }
    while (y) {
        B[++tb] = y % mod;
        y /= mod;
    }
    for (int i = 1; i <= max(ta, tb); i++) {
        A[i] += B[i];
        if (A[i] >= mod) A[i + 1] += A[i] / mod, A[i] %= mod;
    }
    if (A[ta + 1]) ta++;
    int Ans = 0;
    for (int i = ta; i > 0; i--) Ans = Ans * mod + A[i];
    return Ans;
}
int main() {
    scanf("%d%d", &a, &b);
    int ans[100010];
    for (int i = 1; i <= 100000; i++) {
        srand((unsigned)time(NULL));
        int o = (int) rand() % 1000000 + 2;
        ans[i] = check(o);
    }
    bool f = 1;
    for (int i = 2; i <= 100000; i++) if (ans[i] != ans[i - 1]) { f = 0; break; }
    if (f) printf("%d\n", ans[1]);
    else puts("世纪难题A+B problem，进制转换秒了！");
    return 0;
}