HDU6058-Kanade's sum

Kanade's sum

                                                                     Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                             Total Submission(s): 2496    Accepted Submission(s): 1037

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

15 21 2 3 4 5

 

Sample Output

30

 

Source

2017 Multi-University Training Contest - Team 3

 

Recommend

liuyiding

 

题意：给出一个序列，问这个序列中所有区间第k大的为多少

解题思路：用链表来处理， 每次找到链表中最小的数字，那么其他数字必然都是比它大的，处理出它前面的k个数字和后面的k个数字，那么前后延伸的长度就是对该点的贡献，然后求一下乘积的和即可

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;const LL mod=1000000007;int n,k,x[500010],a[100],b[100];;int pos[500010],pre[500010],nt[500010];LL ans;void change(int x){    pre[nt[x]]=pre[x];    nt[pre[x]]=nt[x];}LL solve(int x){    int sum1=0,sum2=0;    for(int i=x; i; i=pre[i])    {        a[++sum1]=i-pre[i];        if(sum1==k) break;    }    for(int i=x; i<=n; i=nt[i])    {        b[++sum2]=nt[i]-i;        if(sum2==k) break;    }    LL sum=0;    for(int i=1; i<=sum1; i++)        if(k-i+1<=sum2) sum+=1LL*a[i]*b[k-i+1];    return sum;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&k);        for(int i=1; i<=n; i++)scanf("%d",&a[i]),pos[a[i]]=i;        for(int i=0; i<=n+1; i++) pre[i]=i-1,nt[i]=i+1;        pre[0]=0;        nt[n+1]=n+1;        ans=0;        for(int i=1; i<=n; i++)        {            int x=pos[i];            ans+=solve(x)*i;            change(x);        }        printf("%lld\n",ans);    }    return 0;}