AT_arc168_e [ARC168E] Subsegments with Large Sums 题解

套路地考虑 wqs 二分，记 $f(x)$ 表示分成 $x$ 段，$\ge s$ 的段数的最大值。

但是发现，$\ge s$ 的段可以拼上旁边一个小段，构成一段坐标系上的横条，所以 $f(x)$ 其实并不是凸的。

由上的性质，可以知道，答案关于 $k$ 具有单调性，我们尝试定义 $g(x)$ 表示 $\ge s$ 的段有 $x$ 个的情况下，划分的最大段数。

但是最大段数的计算需要涉及到已选的段长，我们重新定义 $g(x)=\sum_{i=1}^{x} r_i-l_i$ 即 $\ge s$ 的段的长度减一的和的最小值，若 $g(x)\le n-k$，则我们声称可以划分成 $k$ 段。

感性理解起来，$g(x)$ 特别容易凸，于是我们对 $g$ 进行 wqs 二分，dp 记录前缀最小代价及对应下的最小个数，转移考虑从以当前点为右端点的最近合法点转移即可。

时间复杂度 $\mathcal O(n\log^2 n)$。

// LUOGU_RID: 139532006
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
//#pragma GCC optimize("Ofast,fast-math")
//#pragma GCC target("avx,avx2")
//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
 #define int long long
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef double db;
#define F(i, a, b) for(int i = a; i <= (b); ++i)
#define F2(i, a, b) for(int i = a; i < (b); ++i)
#define dF(i, a, b) for(int i = a; i >= (b); --i)
template<typename T> void debug(string s, T x) {
	cerr << "[" << s << "] = [" << x << "]\n";
}
template<typename T, typename... Args> void debug(string s, T x, Args... args) {
	for (int i = 0, b = 0; i < (int)s.size(); i++) if (s[i] == '(' || s[i] == '{') b++;
	else if (s[i] == ')' || s[i] == '}') b--;
	else if (s[i] == ',' && b == 0) {
		cerr << "[" << s.substr(0, i) << "] = [" << x << "] | ";
		debug(s.substr(s.find_first_not_of(' ', i + 1)), args...);
		break;
	}
}
#ifdef ONLINE_JUDGE
#define Debug(...)
#else
#define Debug(...) debug(#__VA_ARGS__, __VA_ARGS__)
#endif
#define pb push_back
#define fi first
#define se second
#define Mry fprintf(stderr, "%.3lf MB\n", (&Med - &Mbe) / 1048576.0)
#define Try cerr << 1e3 * clock() / CLOCKS_PER_SEC << " ms\n";
typedef long long ll;
// namespace Fread {const int SIZE = 1 << 17; char buf[SIZE], *S, *T; inline char getchar() {if (S == T) {T = (S = buf) + fread(buf, 1, SIZE, stdin); if (S == T) return '\n';} return *S++;}}
// namespace Fwrite {const int SIZE = 1 << 17; char buf[SIZE], *S = buf, *T = buf + SIZE; inline void flush() {fwrite(buf, 1, S - buf, stdout), S = buf;} inline void putchar(char c) {*S++ = c;if (S == T) flush();} struct NTR {~NTR() {flush();}} ztr;}
// #ifdef ONLINE_JUDGE
// #define getchar Fread::getchar
// #define putchar Fwrite::putchar
// #endif
inline int ri() {
	int x = 0;
	bool t = 0;
	char c = getchar();
	while (c < '0' || c > '9') t |= c == '-', c = getchar();
	while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
	return t ? -x : x;
}
inline void wi(int x) {
	if (x < 0) {
		putchar('-'), x = -x;
	}
	if (x > 9) wi(x / 10);
	putchar(x % 10 + 48);
}
inline void wi(int x, char s) {
	wi(x), putchar(s);
}
bool Mbe;
// mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
const int mod = 998244353;
const int inf = 0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3f;
const int _ = 3e5 + 5;

int n, k, s, a[_], sum[_], pre[_];

pii f[_];

void work(int x) {
	F(i, 1, n) {
		f[i] = f[i - 1];
		if(pre[i]) f[i] = min(f[i], make_pair(f[pre[i] - 1].fi + (i - pre[i]) - x, f[pre[i] - 1].se + 1));
	}
}

bool chk(int lim) {
	int l = 1, r = n, mid, res = 0;
	while(l <= r) {
		mid = (l + r) >> 1;
		work(mid);
		if(f[n].se <= lim) l = mid + 1, res = mid;
		else r = mid - 1;
	}
	work(res);
	return f[n].fi + res * lim <= n - k;
}

bool Med;
signed main() {
	// Mry;
	n = ri(), k = ri(), s = ri();
	F(i, 1, n) a[i] = ri(), sum[i] = sum[i - 1] + a[i];
	int l = 0;
	F(i, 1, n) {
		while(sum[i] - sum[l] >= s) l++;
		pre[i] = l;
	}
	l = 1;
	int r = k, ans = 0;
	while(l <= r) {
		int mid = (l + r) >> 1;
		if(chk(mid)) l = mid + 1, ans = mid;
		else r = mid - 1;
	}
	cout << ans << '\n';
	// Try;
	return 0;
}