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/*
Runda 5 - 
*/

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<cmath>
#include<queue>
#include<queue>
#include<stack>
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#include <chrono>

using namespace std;
using namespace __gnu_pbds;

typedef vector<int> VI;
typedef vector<VI> VVI;
typedef long long LL;
typedef unsigned long long ULL;
typedef long double LD;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef vector<LL> VLL;
typedef vector<LD> VLD;
typedef vector<VLL > VVLL;
typedef vector<VLD > VVLD;
typedef vector<PII > VPII;
typedef vector<PLL > VPLL;
template<class T> using min_heap = priority_queue<T, vector<T>, greater<T>>;

#define FOR(x, b, e) for(int x=b; x<=(e); ++x)
#define FORD(x, b, e) for(int x=b; x>=(e); --x)
#define REP(x, n) for(int x=0; x<(n); ++x)
#define VAR(v, n) __typeof(n) v = (n)
#define ALL(c) (c).begin(), (c).end()
#define SIZE(x) ((int)(x).size())
#define FOREACH(i, c) for(VAR(i, (c).begin()); i != (c).end(); ++i)
#define PB push_back
#define ST first
#define ND second
#define THIS (*this)
#define LSB(x) (x & -x)
#define SQR(x) ((x)*(x))

bool possible(int k, VI &a) {
    VI ends(a.size() + 1, 0);
    int current = 0;
    REP(i, a.size()) {
        current -= ends[i];
        if (i <= a.size() - k) {
            int diff = a[i] - current;
            if (diff < 0) return false;
            current += diff;
            if (i + k <= a.size()) ends[i + k] += diff;
        } else if (current != a[i]) return false;
    }
    return true;
}

int main() {
	ios_base::sync_with_stdio(0);
    cin.tie(NULL); cout.tie(NULL);
    int n; cin >> n;
    VI a(n);
    LL sum = 0;
    REP(i, n) {
        cin >> a[i];
        sum += a[i];
    }
    VI div; div.reserve(n);
    for (LL d = 1; d * d <= sum; d++) if (sum % d == 0) {
        if (d <= n) div.PB(d);
        if (d * d != sum && sum / d <= n) div.PB(sum / d);
    }
    sort(div.rbegin(), div.rend());
    REP(i, div.size()) if (possible(div[i], a)) {
        cout << div[i] << '\n';
        return 0;
    }

    return 0;
}