題目

題目連結 --> https://codeforces.com/contest/1358/problem/D

思路及程式碼實現

使勁想可以想到，在選擇一端連續的區間時，最划算的選擇應該是：

• 左端點選擇某個月的某個節點，右端點選擇某個月的月末

另外

• 由於可能跨年份，我們先把兩年拼接在一起

• 接著列舉每一個右端點

• 二分查詢滿足題意的左端點 -- 即找到一個最小的L,使得 sum [ r ] - sum [ l ] <= x

c++程式碼實現

#include <algorithm>
#include <cmath>
#include <cstring>
#include <deque>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <vector>
#define all(x) x.begin(), x.end()
#define lowbit(x) x&(-x)
#define pb push_back
#define int long long
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
typedef pair<int, int> PII;

const int N = 4e5 + 20;
// int e[N], ne[N], h[N], idx;
// char mp[N][N];
int n, x, num[N], a[N], b[N], s[N], c[N];

void solve() {
    cin >> n >> x;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        b[i] = (1 + a[i]) * a[i] / 2;
        s[i] = s[i - 1] + a[i];      //天數字首和
        c[i] = c[i - 1] + b[i];      //代價字首和
    }

    for (int i = 1; i <= n; i++) {
        a[n + i] = a[i], b[n + i] = b[i];
        s[n + i] = s[n + i - 1] + a[n + i];
        c[n + i] = c[n + i - 1] + b[n + i];
    }

    n *= 2;
    int ans = -1;

    for (int i = 1; i <= n; i++) {
        if (s[i] < x)
            continue;
        int p = lower_bound(s + 1, s + n + 1, s[i] - x) - s;
        int tmp = c[i] - c[p];
        int len = s[p] + x - s[i];
        tmp += (a[p] + a[p] - len + 1) * len / 2;

        ans = max(ans, tmp);
    }
    cout << ans << endl;
}

signed main() {
    int tt = 1;
    // cin >> tt;
    while (tt--)
        solve();
    return 0;
}