题目传送门
https://lydsy.com/JudgeOnline/problem.php?id=2600
题解
随便写一个比较简单的 two pointers 练习题。
首先答案肯定是一个原序列上的区间。然后一个区间的费用——显然选择的点应该是这个区间的中位数（初中数学）。
然后一个区间的费用肯定比其子区间费用要大。
所以可以直接 two-pointers 来做。
时间复杂度 \(O(n)\)。
#include<bits/stdc++.h>#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_backtemplate<typename A, typename B> inline char smax(A &a, const B &b) {return a < b ? a = b, 1 : 0;}
template<typename A, typename B> inline char smin(A &a, const B &b) {return b < a ? a = b, 1 : 0;}typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;template<typename I> inline void read(I &x) {
int f = 0, c;
while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
x = c & 15;
while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
f ? x = -x : 0;
}const int N = 100000 + 7;int n, m;
ll b;
int a[N];
ll s[N];inline ll get(int l, int r) {
int mid = (l + r) >> 1;
return s[r] - s[mid] - (ll)(r - mid) * a[mid] + (ll)(mid - l) * a[mid] - (s[mid - 1] - s[l - 1]);
}inline void work() {
int p = 1, ans = 0;
for (int i = 1; i <= n; ++i) {
smax(p, i);
while (p < n && get(i, p + 1) <= b) ++p;
smax(ans, p - i + 1);
}
printf("%d\n", ans);
}inline void init() {
read(n), read(m), read(b);
for (int i = 1; i <= n; ++i) read(a[i]), s[i] = s[i - 1] + a[i];
}int main() {
#ifdef hzhkk
freopen("hkk.in", "r", stdin);
#endif
init();
work();
fclose(stdin), fclose(stdout);
return 0;
}
