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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