单调性显然,双端队列队列维护严格单调递减手写双端队列真的可恶。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int a = (b); a <= (c); ++ a)
#define nR(a,b,c) for(register int a = (b); a >= (c); -- a)
#define Max(a,b) ((a) > (b) ? (a) : (b))
#define Min(a,b) ((a) < (b) ? (a) : (b))
#define Fill(a,b) memset(a, b, sizeof(a))
#define Abs(a) ((a) < 0 ? -(a) : (a))
#define Swap(a,b) a^=b^=a^=b
#define ll long long
#define ON_DEBUG
#ifdef ON_DEBUG
#define D_e_Line printf("
----------
")
#define D_e(x) cout << #x << " = " << x << endl
#define Pause() system("pause")
#define FileOpen() freopen("in.txt","r",stdin);
#else
#define D_e_Line ;
#define D_e(x) ;
#define Pause() ;
#define FileOpen() ;
#endif
struct ios{
template<typename ATP>ios& operator >> (ATP &x){
x = 0; int f = 1; char c;
for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-') f = -1;
while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
x*= f;
return *this;
}
}io;
using namespace std;
//#include <queue>
const int N = 2000007;
int a[N];
long long sum[N], x[N];
//deque<int> q;
int q[N], t, h = 1;
int main(){
//FileOpen();
int n, d;
long long Val;
io >> n >> Val >> d;
R(i,1,n){
io >> a[i];
sum[i] = sum[i - 1] + a[i];
}
R(i,d,n){
x[i] = sum[i] - sum[i - d];
}
int ans = d;
q[++t] = d;
int l = 1;
R(r,d,n){
while(h <= t && x[r] >= x[q[t]]) --t;
q[++t] = r;
while(h <= t && q[h] - d + 1< l) ++h;
while(h <= t && sum[r] - sum[l - 1] - x[q[h]] > Val){
++l;
while(h <= t && q[h] - d + 1 < l) ++h;
}
ans = Max(ans, r - l + 1);
}
printf("%d", ans);
return 0;
}