HDU 6058 Kanade's sum —— 2017 Multi-University Training 3

Kanade's sum

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2512    Accepted Submission(s): 1045

Problem Description

Give you an array A[1..n]of length n. 

Let f(l,r,k) be the k-th largest element of A[l..r].

Specially , f(l,r,k)=0 if r−l+1<k.

Give you k , you need to calculate ∑nl=1∑nr=lf(l,r,k)

There are T test cases.

1≤T≤10

k≤min(n,80)

A[1..n] is a permutation of [1..n]

∑n≤5∗105

 

Input

There is only one integer T on first line.

For each test case,there are only two integers n,k on first line,and the second line consists of n integers which means the array A[1..n]

 

Output

For each test case,output an integer, which means the answer.

 

Sample Input

1

5 2

1 2 3 4 5

 

Sample Output

30

 

题目大意：给定数列A，A是1,2,...,n的一个排序，求数列中所有区间的第k小的数之和。

思路：对于1~n的某个数 i，向左向右分别数出 k 个比 i 大的数，求得第 k 小的数是 i 的最大区间。然后在这个最大区间里，从左往右每次取k个不小于 i 的数，每一次更新区间时加上前后两个区间的差值即可。

 

AC代码（借鉴了下网上的代码）：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<fstream> 
using namespace std;
int a[500005];
long long l[90], r[90];
int main()
{
    int T,n,k;
    //ifstream cin("ylq.txt");
    cin>>T; 
    while(T--)
    {
        scanf("%d %d", &n, &k);
        //cin>>n>>k;
        for(int i=1;i<=n;i++) scanf("%d", &a[i]);
        int t;
        long long ans=0;
        for(int i=1;i<=n;i++){
            t=0;
            l[t++]=i;
            int j;
            for(j=i-1;j>0&&t<k;j--){
                if(a[j]>a[i]){
                    l[t++]=j;
                }
            }
            long long sum=0;
            if(t==k)
            {
                int tmp=1;
                for(;j>0;j--){
                    if(a[j]<a[i]) tmp++;
                    else break;
                }
                sum+=tmp;
                for(j=i+1;j<=n&&t>=0;j++){
                    if(a[j]<a[i]) sum+=tmp;
                    else{
                        t--;
                        if(t==0) break;
                        tmp=l[t-1]-l[t];
                        sum+=tmp;
                    }
                }
            }
            else
            {
                for(j=i+1;j<=n&&t<k;j++){
                    if(a[j]>a[i]) l[t++]=j;
                }
                if(t==k)
                {
                    sort(l, l+t);
                    int tmp=l[0];
                    int p=0;
                    sum+=tmp;
                    for(;j<=n&&p<t;j++){
                        if(a[j]<a[i]) sum+=tmp;
                        else{
                            p++;
                            if(l[p]>i) break;    
                            tmp=abs(l[p]-l[p-1]);
                            sum+=tmp;
                        }
                        
                    }
                }
            }
            ans+=(long long)(sum*a[i]);
        } 
        cout<<ans<<endl;
    }
}