2017 Multi-University Training Contest

Kanade's sum

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2109    Accepted Submission(s): 862

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

 
一开始先维护一个满的链表，然后从小到大删除，每次算完一个数，就在链表里面删除，算x的时候，保证删除的数都比x小，都可以用来算贡献。i和pre[i]和nxt[i]的距离就是小于当前的数的数目+1

#include <bits/stdc++.h>using namespace std;const int N = 5e5 + 10;typedef long long LL;int pre[N], nxt[N], v[N], pos[N], n, k;LL a[N], b[N];LL solve(int x){    int c1 = 0, c2 = 0;    for (int i = x; i && c1 <= k; i = pre[i])    {        a[++c1] = i - pre[i];    }    for (int i = x; i <= n && c2 <= k; i = nxt[i])    {        b[++c2] = nxt[i] - i;    }    LL ans = 0;    for (int i = 1; i <= c1; i++)        if (k - i + 1 <= c2 && k - i + 1 >= 1)        {            ans += a[i] * b[k - i + 1];        }    return ans;}void del(int x){    pre[nxt[x]] = pre[x];    nxt[pre[x]] = nxt[x];}int main(){    int t;    scanf("%d", &t);    while (t--)    {        scanf("%d%d", &n, &k);        for (int i = 1; i <= n; i++)        {            scanf("%d", &v[i]), pos[v[i]] = i;        }        for (int i = 0; i <= n + 1; i++)        {            pre[i] = i - 1, nxt[i] = i + 1;        }        pre[0] = 0;        nxt[n + 1] = n + 1;        LL ans = 0;        for (int i = 1; i <= n; i++)        {            ans += solve(pos[i]) * i;            del(pos[i]);        }        printf("%lld\n", ans);    }    return 0;}