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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#794209#5417. Chat ProgramprokingTL 20ms5132kbC++142.5kb2024-11-30 12:54:462024-11-30 12:54:46

Judging History

你现在查看的是最新测评结果

  • [2024-11-30 12:54:46]
  • 评测
  • 测评结果:TL
  • 用时:20ms
  • 内存:5132kb
  • [2024-11-30 12:54:46]
  • 提交

answer

#include <bits/stdc++.h>

#define F(i, a, b) for (int i = a; i <= b; i ++)
#define G(i, a, b) for (int i = a; i >= b; i --)
#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) < (b) ? (a) : (b))
#define mx(a, b) ((a) = max(a, b))
#define mn(a, b) ((a) = min(a, b))
#define get getchar()
#define putc putchar
#define outarr(a, L, R) { prllf(#a"[%d..%d]", L, R); F(i, L, R) W(a[i]), putc(' ') ; putc('\n'); }
#define out1(x) { printf(#x" = "), W(x), putc('\n'); }
#define out2(x, y) { printf(#x" = "), W(x), printf(" "#y" = "), W(y), putc('\n'); }
#define out3(x, y, z) { printf(#x" = "), W(x), printf(" "#y" = "), W(y), printf(" "#z" = "), W(z), putc('\n'); }
#define out4(x, y, z, w) { printf(#x" = %d", x); printf(" "#y" = %d", y); printf(" "#z" = %d", z), printf(" "#w" = %d", w), putc('\n'); }
#define Assert(x, y) { if (!(x)) { printf("Wrong %d", y); exit(0); } }
#define mem(a, b) memset(a, b, sizeof a)
#define mec(a, b) memcpy(a, b, sizeof a)
#define pi pair<int, int>
#define V vector <int>
#define ll long long
#define get getchar()
#define pb push_back
#define putc putchar
#define pb push_back
#define db double
#define fi first
#define se second

using namespace std;

template <typename Int>
void R(Int &x) {
	char c = get; x = 0; Int t = 1;
	for (; !isdigit(c); c = get) t = (c == '-' ? -1 : t);
	for (; isdigit(c); x = (x << 3) + (x << 1) + c - '0', c = get); x *= t;
}
template <typename Int>
void W(Int x) {
	if (x < 0) { putc('-'); x = - x; }
	if (x > 9) W(x / 10); putc(x % 10 + '0');
}

using namespace std;

const int N = 2e5 + 10;
int n, k, m, c, d, a[N], f[N];

bool ok(int p) {
//	out1(p);
	F(i, 1, n) f[i] = 0;
	int Max = 0, cnt = 0;
	F(i, 1, n) {
//		out2(a[i], p);
		if (a[i] >= p) {
			cnt ++;
			continue;
		}
//		out1(d);
		if (d == 0) {
			if (c + a[i] >= p) {
				if (i + m - 1 <= n)
					f[i] ++, f[i + 1] --;
			}
			continue;
		}
		int w = a[i] + c >= p ? 0 : ((p - c - a[i] - 1) / d + 1);
//		if (p == 6) {
//			out4(p - c - a[i] - 1, d, w, a[i]);
//			out2(max(1, i - m + 1), max(1, i - w + 1));
//			putc('\n');
//		}
		if (w <= m)
			f[max(1, i - m + 1)] ++, f[max(1, i - w + 1)] --;
	}
	F(i, 1, n) f[i] += f[i - 1], mx(Max, cnt + f[i]);
//	out2(cnt, Max);
	return Max >= k;
}

int main() {
//	freopen("data.in","r",stdin);

	R(n), R(k), R(m), R(c), R(d);
	F(i, 1, n) R(a[i]);

	int l = 0, r = 2e9, ans = -1;
	while (l <= r) {
		int md = l + r >> 1;
		if (ok(md))
			ans = md, l = md + 1;
		else
			r = md - 1;
	}
	
	W(ans);
}

詳細信息

Test #1:

score: 100
Accepted
time: 0ms
memory: 3580kb

input:

6 4 3 1 2
1 1 4 5 1 4

output:

4

result:

ok 1 number(s): "4"

Test #2:

score: 0
Accepted
time: 0ms
memory: 3636kb

input:

7 3 2 4 0
1 9 1 9 8 1 0

output:

9

result:

ok 1 number(s): "9"

Test #3:

score: 0
Accepted
time: 0ms
memory: 3640kb

input:

8 3 5 0 0
2 0 2 2 1 2 1 8

output:

2

result:

ok 1 number(s): "2"

Test #4:

score: 0
Accepted
time: 20ms
memory: 5132kb

input:

200000 200000 100000 0 1000000000
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ...

output:

0

result:

ok 1 number(s): "0"

Test #5:

score: -100
Time Limit Exceeded

input:

200000 1 100000 1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 100000000...

output:


result: