2017 Multi-University Training Contest - Team 3 Kanade's sum hd6058
地址:http://acm.split.hdu.edu.cn/showproblem.php?pid=6058
题目:
Kanade's sum
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 505 Accepted Submission(s): 176
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
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]
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
思路:
很容易想到按公式算是不可行的(O(n^2)的时间复杂度),然后想到枚举计算每个数的贡献:即第k大为x的区间个数乘以x
然后只要考虑怎么快速求出第k大为x的区间个数。如果能知道左边大于x的80个数的位置和右边大于x的80个数的位置就可以计算区间个数了。
之后想到用从小到大枚举或者从大到小,用链表维护即可。
比赛时用的set模拟的,结果被卡了,T的连妈都不认识。还是觉得时限卡太紧了。
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 #define MP make_pair 6 #define PB push_back 7 typedef long long LL; 8 typedef pair<int,int> PII; 9 const double eps=1e-8; 10 const double pi=acos(-1.0); 11 const int K=1e6+7; 12 const int mod=1e9+7; 13 14 int n,k,p[K],pre[K],nxt[K],pos[K],tl[85],tr[85]; 15 LL ans; 16 17 int main(void) 18 { 19 int t;cin>>t; 20 while(t--) 21 { 22 ans=0; 23 scanf("%d%d",&n,&k); 24 for(int i=1;i<=n;i++) 25 scanf("%d",p+i),pos[p[i]]=i; 26 for(int i=1;i<=n;i++) 27 pre[i]=i-1,nxt[i]=i+1; 28 pre[1]=0,nxt[n]=n+1; 29 for(int i=1;i<=n;i++) 30 { 31 int la=0,lb=0; 32 for(int j=pos[i];j>0&&la<=k;j=pre[j]) 33 tl[la++]=j-pre[j]; 34 for(int j=pos[i];j<=n&&lb<=k;j=nxt[j]) 35 tr[lb++]=nxt[j]-j; 36 for(int j=0;j<la;j++) 37 if(k-j-1<lb) 38 ans+=i*1LL*tl[j]*tr[k-j-1]; 39 pre[nxt[pos[i]]]=pre[pos[i]]; 40 nxt[pre[pos[i]]]=nxt[pos[i]]; 41 } 42 printf("%lld\n",ans); 43 } 44 return 0; 45 }
作者:weeping
出处:www.cnblogs.com/weeping/
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