Submission #6502541

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Source Code Expand

#include<cmath>
#include<cstdio>
#include<iostream>
using namespace std;
typedef long long ll;
ll s, n, m;
ll f(ll b, ll n) {
    return n < b ? n : f(b, n / b) + n % b;
}
int main() {
    cin >> n >> s; m = sqrt(n) + 1;
    if (s > n) return puts("-1"), 0;
    if (s == n) return printf("%lld\n", s + 1), 0;
    for (ll i = 2; i <= m; ++i)
        if (f(i, n) == s) return printf("%lld\n", i), 0;
    ll b, q, res = 1e11; n -= s;
    for (ll p = 1; p * p <= n; ++p) {
        if (n % p == 0 ) {
            b = n / p + 1; q = s - p;
            if (q >= 0 && b >= 2 && q < b && p < b)
                res = min(res, b);
        }
    } printf("%lld\n", res == 1e11 ? -1 : res);
    return 0;
}

Submission Info

Judge Result