题解 | #相邻的糖果#

贪心 + 双指针

注意到删除后面的糖果数量一定比删除前面的好, 因此贪心的删除后面的糖果, 并且维护当前区间是合法的

算法时间复杂度

#include <bits/stdc++.h>

#define x first
#define y second
#define all(x) x.begin(), x.end()

using namespace std;
using i128 = __int128;
using u128 = unsigned __int128;

typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;

const int N = 1e5 + 10, MOD = 998244353;
const int INF = 1e9;
const LL LL_INF = 1e18;
const LD EPS = 1e-8;
const int dx4[] = {-1, 0, 1, 0}, dy4[] = {0, 1, 0, -1};
const int dx8[] = {-1, -1, -1, 0, 0, 1, 1, 1}, dy8[] = {-1, 0, 1, -1, 1, -1, 0, 1};

istream &operator>>(istream &is, i128 &val) {
    string str;
    is >> str;
    val = 0;
    bool flag = false;
    if (str[0] == '-') flag = true, str = str.substr(1);
    for (char &c: str) val = val * 10 + c - '0';
    if (flag) val = -val;
    return is;
}

ostream &operator<<(ostream &os, i128 val) {
    if (val < 0) os << "-", val = -val;
    if (val > 9) os << val / 10;
    os << static_cast<char>(val % 10 + '0');
    return os;
}

void solve() {
    int n, m, x;
    cin >> n >> m >> x;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; ++i) cin >> a[i];
    int l = 1;
    LL cur = 0;
    LL ans = 0;

    for (int r = 1; r <= n; ++r) {
        if (r > m) {
            cur -= a[l];
            l++;
        }
        cur += a[r];
        if (cur > x) {
            LL d = cur - x;
            ans += d;
            a[r] -= d;
            cur = x;
        }
    }

    cout << ans << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int T = 1;
    while (T--) solve();
    cout << fixed << setprecision(15);

    return 0;
}