AT4842 [ABC136E] Max GCD

原题链接

• 题解：解决了历史遗留问题。很显然的是，这样操作总和不会变，然后所有 (gcd) 的可能的取值都是会是 (sum) 的因子，所以可以 (O(sqrt {sum})) 的时间复杂度来枚举 (gcd)，然后 (O(n)) 遍历用最小操作来看是否是小于 (k)，最后取最大。
• 代码：

#include <algorithm>
#include <cmath>
#include <cstring>
#include <iostream>
using namespace std;
typedef long long ll;
const ll N = 1e6 + 9, mod = 1e9 + 7;

ll a[N];
ll b[N];
ll sum = 0;
ll n, k;
ll ans;
void check(ll p) {
    ll ssum = 0;
    for (ll i = 1; i <= n; i++) {
        b[i] = a[i] % p;
    }
    sort(b + 1, b + 1 + n);
    ll sub = 0;
    ll add = 0;
    int l = 1, r = n;
    while (l <= r) {
        while (l <= r && b[l] == 0) {
            l++;
        }
        while (l <= r && b[r] == p) {
            r--;
        }
        if (r == l) {
            ssum += min(b[l], p - b[l]);
            break;
        }
        if (l > r) break;
        ssum += b[l];
        ll t = b[l];
        
        b[l] = 0;
        while (l <= r && t > 0) {
            if ( b[r] + t >= p ) {
                t -= (p - b[r]);
                b[r] = p;
                r--;
            } else {
                b[r] += t;
                t = 0;
                break;
            }
        }
    }
    if (ssum <= k) {
        ans = max(ans, p);
    }
}
signed main() {
    cin >> n >> k;
    for (ll i = 1; i <= n; i++) {
        cin >> a[i];
        sum += a[i];
    }
    ans = -1;
    check(sum);
    for (ll i = 1; i * i <= sum; i++) {
        if (sum % i == 0) {
            check(i);
            check(sum / i);
        }
    }
    cout << ans << endl;
    return 0;
}