【题解】CF460C：Present

原题传送门
最小值最大当然是二分答案
然后$O(n)$扫，如果当前数未达到$mid$，就加到$mid$，并且这个数是作为区间左端点的
如何实现区间加法，差分数组就行了
Code：

#include <bits/stdc++.h>
#define maxn 1000010
#define LL long long
using namespace std;
LL a[maxn], d[maxn], n, m, w, ans;

inline int read(){
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

bool check(LL mid){
	int sum = m;
	for (int i = 1; i <= n; ++i) d[i] = 0;
	for (int i = 1; i <= n; ++i){
		d[i] += d[i - 1];
		if (a[i] + d[i] < mid){
			int x = mid - a[i] - d[i];
			if (x > sum) return 0;
			sum -= x, d[i] += x, d[i + w] -= x;
		}
	}
	return 1;
}

int main(){
	n = read(), m = read(), w = read();
	for (int i = 1; i <= n; ++i) a[i] = read();
	LL l = 0, r = 2e9;
	while (l <= r){
		LL mid = (l + r) >> 1;
		if (check(mid)) ans = mid, l = mid + 1; else r = mid - 1;
	}
	printf("%lld\n", ans);
	return 0;
}