题解

$pre_i$ 表示合并前$i$个数的代价
$suf_i$ 表示合并前$n$到$i$个数的代价
枚举1 ~ n求$pre_i+suf_i$最小值就行了

#include <bits/stdc++.h>
#define ll long long
#define endl '\n'
#define pii pair<int, int>
using namespace std;
const int maxn = 2e5 + 5;
int n, k;
int a[maxn];
ll pre[maxn], suf[maxn];
int main() {
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    ll sum = 0;
    for (int i = 1; i <= n; i++) {
        pre[i] = pre[i - 1] + (sum + k - 1) / k;
        sum += a[i];
    }
    sum = 0;
    for (int i = n; i >= 1; i--) {
        suf[i] = suf[i + 1] + (sum + k - 1) / k;
        sum += a[i];
    }
    ll ans = 2e18;
    for (int i = 1; i <= n; i++) {
        ans = min(ans, suf[i] + pre[i]);
    }
    cout << ans << endl;
}