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Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 517 Accepted Submission(s): 250
求有多少个区间,使得区间内的第k大的值>=m.
将不小于m的数看作1,剩下的数看作0,那么只要区间内1的个数不小于k则可行,枚举左端点,右端点可以通过two-pointer求出。
/* *********************************************** Author :guanjun Created Time :2016/8/7 10:02:54 File Name :hdu5806.cpp ************************************************ */ #include <iostream> #include <cstring> #include <cstdlib> #include <stdio.h> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <iomanip> #include <list> #include <deque> #include <stack> #define ull unsigned long long #define ll long long #define mod 90001 #define INF 0x3f3f3f3f #define maxn 10010 #define cle(a) memset(a,0,sizeof(a)) const ull inf = 1LL << 61; const double eps=1e-5; using namespace std; priority_queue<int,vector<int>,greater<int> >pq; struct Node{ int x,y; }; struct cmp{ bool operator()(Node a,Node b){ if(a.x==b.x) return a.y> b.y; return a.x>b.x; } }; bool cmp(int a,int b){ return a>b; } int a[200010]; int sum[200010]; int n,m,k; int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); #endif //freopen("out.txt","w",stdout); int t; cin>>t; while(t--){ scanf("%d%d%d",&n,&m,&k); int x; sum[0]=0; for(int i=1;i<=n;i++){ scanf("%d",&x); if(x>=m)a[i]=1; else a[i]=0; sum[i]=sum[i-1]+a[i]; } ll ans=0; int r=1; for(int l=1;l<=n;l++){//枚举左端点 while(r<=n&&sum[r]-sum[l-1]<k)r++; if(r>n)break; ans+=(n-r+1); } printf("%lld\n",ans); } return 0; }
将不小于m的数看作1,剩下的数看作0,那么只要区间内1的个数不小于k则可行,枚举左端点,右端点可以通过two-pointer求出
HDU5806 NanoApe Loves Sequence Ⅱ
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原文地址:http://www.cnblogs.com/pk28/p/5745762.html