2017CCPC哈尔滨站 K-th Number 【二分】

K-th Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 76    Accepted Submission(s): 26

Problem Description

Alice are given an array A[1..N] with N numbers.

Now Alice want to build an array B by a parameter K as following rules:

Initially, the array B is empty. Consider each interval in array A. If the length of this interval is less than K , then ignore this interval. Otherwise, find the K -th largest number in this interval and add this number into array B .

In fact Alice doesn't care each element in the array B. She only wants to know the M -th largest element in the array B . Please help her to find this number.

 

Input

The first line is the number of test cases.

For each test case, the first line contains three positive numbers N(1≤N≤105),K(1≤K≤N),M . The second line contains N numbers Ai(1≤Ai≤109) .

It's guaranteed that M is not greater than the length of the array B.

 

Output

For each test case, output a single line containing the M -th largest element in the array B .

 

Sample Input

2 5 3 2 2 3 1 5 4 3 3 1 5 8 2

 

Sample Output

3 2

 

题意：给出一个长度为N的数组a[i]，问在这个数组中的所有连续区间中的第K小的数组成的数组b[i]中第M大的数。

思路难得，如果想得到所有区间中的第K小，达到O（n*n）很轻松，转换一下思路，一个区间中，如果第K小的数要大于最终答案，那么这个区间中至少要有K个数大于等于最终答案。我们设最终答案为x，且这个x满足有大于等于m个区间的第k小大于等于x，这个区间的数量我们可以通过枚举每一个区间的端点对于每个左端L，可以找一个最小的r使得,当右端点大于等于r时，[L,r]有k个数大于等于x。所以L为左端点的区间中满足要求的区间数有 n-r+1个，通过区间的数量与m的关系决定二分的走向。

#include
using namespace std;
int n,m,k,a[100001];
long long find(int x)
{
    long long ans=0;
    int q=0;
    int out=0;
    for(int i=1;i<=n;i++){
        while(out=x)
                out++;
        }
        ans+=n-q+1;
        if(a[i]>=x)
            out--;
    }
    return ans;
}
int main(){
    long long m;
    int t,l,r,mid;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d%lld",&n,&k,&m);
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
        }
        l=1,r=1000000000;
        while(l=m)
                l=mid;
            else
                r=mid-1;
        }
        printf("%d\n",r);
    }
    return 0;

}