很神仙的题目啊,考场上只会$n^2$的暴力。。
考虑直接二分一个$mid$,我们来判断最终答案是否可能大于$x$。
判断的时候记录一下前缀最小值即可,
设$s[i]$表示$1-i$中有多少比它大的,要求的长度为$len$,我们记下$s[i - len]$的最小值为$Mi$
若$s[i] - Mi > 0$,那么说明在长度至少为$len$的区间中,大于$mid$的数和小于$mid$的数相互抵消后仍然有比$mid$大的数,此时$mid$是合法的
第一次做这种二分答案,但答案不是给出的数的题。神仙啊qwq
/*
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#include<set>
#include<vector>
//#define int long long
#define LL long long
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
using namespace std;
const int MAXN = 1e5 + 10, INF = 1e9 + 10, mod = 1e9 + 7;
const double eps = 1e-9;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N, len, s[MAXN], a[MAXN];
int check(int x) {
for(int i = 1; i <= N; i++)
if(a[i] < x) s[i] = -1;
else s[i] = 1;//s[i] : 1 - i中有多少比x大的
int Mi = 0;
for(int i = 1; i <= N; i++) {
s[i] += s[i - 1];
if(i >= len) Mi = min(Mi, s[i - len]);
if(i >= len && (s[i] - Mi > 0)) return 1;
}
return 0;
}
int main() {
// freopen("a.in", "r", stdin);
// freopen("c.out", "w", stdout);
N = read(); len = read();
for(int i = 1; i <= N; i++) a[i] = read();
int l = 0, r = 1e9 + 10, ans;
while(l <= r) {
int mid = l + r >> 1;
if(check(mid)) ans = mid, l = mid + 1;//是否有比mid大的解
else r = mid - 1;
}
printf("%d", ans);
return 0;
}
/*
5 4
7 2 3 2 6
*/