## 解法

### 概要

$9$ の倍数判定の一般化である「$n$ の $i$ 進数での数字和は $n \bmod (i - 1)$ に等しい」に気付く。 あとは中国人剰余定理。 一般的なCRTがなくても、互いに素になるようにいくつか数字を選んでやれば十分のはず。 $O(\log n)$ ぐらい。

## 実装

#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; (i) < (int)(n); ++ (i))
#define REP3(i, m, n) for (int i = (m); (i) < (int)(n); ++ (i))
using ll = long long;
using namespace std;

tuple<ll, ll, ll> extgcd(ll a, ll b) {
ll x = 0, y = 1;
for (ll u = 1, v = 0; a; ) {
ll q = b / a;
x -= q * u; swap(x, u);
y -= q * v; swap(y, v);
b -= q * a; swap(b, a);
}
return make_tuple(x, y, b);
}

ll multmod(ll a, ll b, ll m) {
a = (a % m + m) % m;
b = (b % m + m) % m;
ll c = 0;
REP (i, 63) {
if (b & (1ll << i)) {
c += a;
if (c > m) c -= m;
}
a *= 2;
if (a > m) a -= m;
}
return c;
}

pair<ll, ll> crt(pair<ll, ll> eqn1, pair<ll, ll> eqn2) {
ll x1, m1; tie(x1, m1) = eqn1;
ll x2, m2; tie(x2, m2) = eqn2;
if (m1 == 0 or m2 == 0) return make_pair(0ll, 0ll);
assert (1 <= m1 and 1 <= m2);
ll m1_inv, d; tie(m1_inv, ignore, d) = extgcd(m1, m2);
if ((x1 - x2) % d) return make_pair(0ll, 0ll);
ll m = m1 * m2 / d;
ll x = x1 + multmod(multmod(m1 / d, x2 - x1, m), m1_inv, m);
return make_pair((x % m + m) % m, m);
}

int get_a_i(ll n, int i) {
int a_i = 0;
for (; n; n /= i) {
a_i += n % i;
}
return a_i;
}

ll solve(array<int, 31> const & a) {
ll n = 0, m = 1;
REP3 (i, 2, 31) {
tie(n, m) = crt(make_pair(n, m), make_pair(a[i] % (i - 1), i - 1));
if (m > 1e12) break;
}
if (n == 0 or n > 1e12) return -1;
REP3 (i, 2, 31) {
if (get_a_i(n, i) != a[i]) return -1;
}
return n;
}

int main() {
array<int, 31> a;
REP3 (i, 2, 31) cin >> a[i];
ll n = solve(a);
if (n == -1) {
cout << "invalid" << endl;
} else {
cout << n << endl;
}
return 0;
}