NanoApe Loves Sequence Ⅱ

链接：http://acm.split.hdu.edu.cn/showproblem.php?pid=5806

题目：NanoApe, the Retired Dog, has returned back to prepare for for the National Higher Education Entrance Examination!

In math class, NanoApe picked up sequences once again. He wrote down a sequence with  numbers and a number   on the paper.

Now he wants to know the number of continous subsequences of the sequence in such a manner that the -th largest number in the subsequence is no less than  .

Note : The length of the subsequence must be no less than  .

题意：给一串数字，求所有长度大于k，第k大的数字大于等于m的字串个数

分析：这道题很巧妙的用了前缀和来保存从第一个数字到底n个数字里面大于m的数字的个数，之后再用尺取法遍历，记录结果。

题解：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <vector>#include <map>#include <string>#include <cstring>#include <functional>#include <cmath>#include <cctype>#include <cfloat>#include <climits>#include <complex>#include <deque>#include <list>#include <set>#include <utility>using namespace std;int pre[200010];__int64 ans;int n,m,k,temp;int main(){//freopen("in.txt","r",stdin);int T;scanf("%d",&T);while(T--){ans=0;scanf("%d %d %d",&n,&m,&k);for(int i=1;i<=n;i++){scanf("%d",&temp);if(temp>=m)pre[i]=pre[i-1]+1;elsepre[i]=pre[i-1];}int l,r;l=0,r=1;for(;l<n;l++){while(pre[r]-pre[l]<k&&r<=n)r++;if(pre[r]-pre[l]<k)break;ans+=n-r+1;}printf("%I64d\n",ans);}return 0;}