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[];
 long long l[], r[];
 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=;i<=n;i++) scanf("%d", &a[i]);
         int t;
         long long ans=;
         for(int i=;i<=n;i++){
             t=;
             l[t++]=i;
             int j;
             for(j=i-;j>&&t<k;j--){
                 if(a[j]>a[i]){
                     l[t++]=j;
                 }
             }
             long long sum=;
             if(t==k)
             {
                 int tmp=;
                 for(;j>;j--){
                     if(a[j]<a[i]) tmp++;
                     else break;
                 }
                 sum+=tmp;
                 for(j=i+;j<=n&&t>=;j++){
                     if(a[j]<a[i]) sum+=tmp;
                     else{
                         t--;
                         if(t==) break;
                         tmp=l[t-]-l[t];
                         sum+=tmp;
                     }
                 }
             }
             else
             {
                 for(j=i+;j<=n&&t<k;j++){
                     if(a[j]>a[i]) l[t++]=j;
                 }
                 if(t==k)
                 {
                     sort(l, l+t);
                     int tmp=l[];
                     int p=;
                     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-]);
                             sum+=tmp;
                         }

                     }
                 }
             }
             ans+=(long long)(sum*a[i]);
         }
         cout<<ans<<endl;
     }
 }