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题目链接:
Time Limit: 4000/2000 MS (Java/Others)
Memory Limit: 262144/131072 K (Java/Others)
/************************************************
┆ ┏┓ ┏┓ ┆
┆┏┛┻━━━┛┻┓ ┆
┆┃ ┃ ┆
┆┃ ━ ┃ ┆
┆┃ ┳┛ ┗┳ ┃ ┆
┆┃ ┃ ┆
┆┃ ┻ ┃ ┆
┆┗━┓ ┏━┛ ┆
┆ ┃ ┃ ┆
┆ ┃ ┗━━━┓ ┆
┆ ┃ AC代马 ┣┓┆
┆ ┃ ┏┛┆
┆ ┗┓┓┏━┳┓┏┛ ┆
┆ ┃┫┫ ┃┫┫ ┆
┆ ┗┻┛ ┗┻┛ ┆
************************************************ */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
//#include <bits/stdc++.h>
#include <stack>
#include <map>
using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar());
for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + ‘0‘);
putchar(‘\n‘);
}
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=2e5+10;
const int maxn=2e3+14;
const double eps=1e-12;
int n,m,a[N],sum[N],k;
int main()
{
int t;
read(t);
while(t--)
{
read(n);read(m);read(k);
For(i,1,n)
{
read(a[i]);
if(a[i]>=m)sum[i]=sum[i-1]+1;
else sum[i]=sum[i-1];
}
LL ans=0;
int r=1;
For(i,1,n-k+1)
{
r=max(i+k-1,r);
while(sum[r]-sum[i-1]<k&&r<=n)r++;
if(r<=n&&r-i+1>=k)ans=ans+(n-r+1);
}
cout<<ans<<"\n";
}
return 0;
}
hdu-5806 NanoApe Loves Sequence Ⅱ(尺取法)
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原文地址:http://www.cnblogs.com/zhangchengc919/p/5745171.html