Codeforces Round #670 (Div. 2) D - Three Sequences （差分+思维）

显然，为了让 max(b[n],  c[1]) 最小，当 a[i] > a[i - 1]时，b[i] = b[i - 1] + a[i] - a[i - 1], c[i] = c[i - 1]。当a[i] < a[i - 1]时，b[i] = b[i - 1], c[i] = c[i - 1] + a[i] - a[i - 1]

假设K = Σ max(0, a[i] - a[i - 1])，则a[1] - c[1] + K = b[n], 为了最小化b[n]和c[1]，显然c[1] = (a[1] + K) / 2， 每次区间改变都可以用O(1)复杂度维护

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
long long a[N], d[N];
int main() {
    int n;
    scanf("%d", &n);
    long long k = 0;
    for (int i = 1; i <= n; i++) {
        scanf("%lld", &a[i]);
        d[i] =  a[i] - a[i - 1];
        if (i != 1) k = k + max(0ll, d[i]);
    }
    printf("%lld\n", max(a[1] + k - (a[1] + k) / 2, (a[1] + k) / 2));
    int q;
    scanf("%d", &q);
    while (q--) {
        long long l, r, x;
        scanf("%lld %lld %lld", &l, &r, &x);
        if (l == 1)    a[1] += x;
        else {
            long long t = d[l] + x;
            k += max(t, 0ll) - max(d[l], 0ll);
            d[l] += x;
        }
        if (r != n) {
            long long t = d[r + 1] - x;
            k += max(t, 0ll) - max(d[r + 1], 0ll);
            d[r + 1] -= x;
        }
        long long X = (a[1] + k) / 2;
        printf("%lld\n", max(X, a[1] + k - X));
    }
    return 0;
}