二分

挺容易发现的二分思路，单调性比较明显，而且check函数直接把大于等于答案的累加进去，求解判断下就可以了。
注意特判个k=1，就行了。

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cstring>
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(vv) (vv).begin(), (vv).end()
#define endl "\n"
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 1e6 + 10;
ll a[N], n, k;

bool check(ll x) {
    ll ans = 0;
    for (int i = 1; i <= n; i++)
        if (a[i] > x)    ans += (a[i] - x + k - 2) / (k - 1);
    return ans <= x;
}

int main() {
    while (scanf("%lld", &n) != EOF) {
        // cout << n << endl;
        for (int i = 1; i <= n; i++)
            scanf("%lld", &a[i]);
        ll l = 1, r = 1e9;
        scanf("%lld", &k);
        if (k == 1) {
            printf("%d\n", *max_element(a + 1, a + 1 + n));
            continue;
        }
        while (l < r) {
            ll mid = l + r >> 1;
            if (check(mid))  r = mid;
            else l = mid + 1;
        }
        printf("%lld\n", l);
        // puts("");
    }
    return 0;
}