HDU 6058 Kanade's sum（思维）

Kanade's sum

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

Source

2017 Multi-University Training Contest - Team 3

题意：给出长为n的序列a，给出整数k，求对于序列a，每个区间第k大的数之和

题解：相当于求每个数字分别是多少个区间的第k大。记录每个数出现的位置，维护一个set（相当于链表），从大到小遍历，将每个数的位置插入set；对于每个数，只需在其两边找k个比它大的数，对起始位置进行遍历，计算每个数对答案的贡献，求和。

#include<iostream>#include<stdio.h>#include<string.h>#include<vector>#include<algorithm>#include<math.h>#include<queue>#define INF 1e9#define ll long long#define maxn 300005#include<set>#include<stack>using namespace std;int a[500005];int pos[500005];int ri[500005];int le[500005];int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,k;        memset(a,0,sizeof(a));        memset(pos,0,sizeof(pos));        scanf("%d%d",&n,&k);        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            pos[a[i]]=i;            le[i]=0;            ri[i]=n+1;        }        set<int>s;        set<int>::iterator it;        s.insert(0);        s.insert(n+1);        ll ans=0;        le[n+1]=0;        ri[n+1]=n+1;        ri[0]=n+1;        le[0]=0;        for(int x=n;x>=1;x--)        {            s.insert(pos[x]);            it=s.find(pos[x]);            it++;            int p=*it;            int pp=le[p];            le[pos[x]]=pp;            ri[pos[x]]=p;            le[p]=pos[x];            ri[pp]=pos[x];            int l=pos[x];            for(int j=0;j<k&&l!=0;j++)                l=le[l];            int r=l;            for(int j=0;j<k&&r!=n+1;j++)                r=ri[r];            int nowl=l;            int nowr=r;            for(int j=0;j<k;j++)            {                if(nowl==pos[x]||nowr==n+1)                    break;                int nextl=ri[nowl];                int nextr=ri[nowr];                ans+=(1ll*(nextl-nowl)*(nextr-nowr)*x);             //   cout<<ans<<endl;                nowl=nextl;                nowr=nextr;            }        }       printf("%lld\n",ans);    }    return 0;}