HDU多校1003

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

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

Specially , f(l,r,k)=0f(l,r,k)=0 if r−l+1<kr−l+1<k.

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

There are T test cases.

1≤T≤101≤T≤10

k≤min(n,80)k≤min(n,80)

A[1..n] is a permutation of [1..n]A[1..n] is a permutation of [1..n]

∑n≤5∗105∑n≤5∗105
Input
There is only one integer T on first line.

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

题意：求所有区间第k大值的和

思路：链表，从依次最小数到第k大数处理，处理一个删一个，删除之后会留下一个空位，处理一个数只需得到大于此数的左边k个数和大于此数的右边k个数，上代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;const ll N=500005;ll pos[N],pre[N],nex[N];ll lef[105],righ[105];void shanchu(ll x){    ll p=pos[x],l=pre[p],r=nex[p];    nex[l]=r;    pre[r]=l;}int main(){    ll T;    scanf("%lld",&T);    while(T--){        ll n,k,a;        scanf("%lld%lld",&n,&k);        for(ll i=1;i<=n;i++){            scanf("%lld",&a);            pos[a]=i;        }        for(ll i=1;i<=n;i++){            pre[i]=i-1;            nex[i]=i+1;        }        ll ans=0;        for(ll x=1;x<=n;x++){            ll p=pos[x];            ll l_num=0,r_num=0;            for(ll d=p;d!=0&&l_num<=k;d=pre[d])lef[++l_num]=d;            for(ll d=p;d!=n+1&&r_num<=k;d=nex[d])righ[++r_num]=d;            lef[++l_num]=0;            righ[++r_num]=n+1;            for(ll i=1;i<l_num;i++){                if(k+1-i>=1&&k+1-i<=r_num-1){                    ans+=(righ[k+1-i+1]-righ[k+1-i])*(lef[i]-lef[i+1])*x;                }            }            shanchu(x);        }        printf("%lld\n",ans);    }    return 0;}