Codeforces86DPowerfularray（莫队）

An array of positive integers a1, a2, ..., an is given. Let us consider its arbitrary subarray al, al + 1..., ar, where 1 ≤ l ≤ r ≤ n. For every positive integer s denote by Ks the number of occurrences of s into the subarray. We call the power of the subarray the sum of products Ks·Ks·s for every positive integer s. The sum contains only finite number of nonzero summands as the number of different values in the array is indeed finite.

You should calculate the power of t given subarrays.

 1 #include 
 2 #include 
 3 #include 
 4 using namespace std;
 5 
 6 typedef long long LL;
 7 const int N=1e6+10;
 8 struct node{
 9     int l,r,id;
10 }Q[N];
11 
12 LL a[N],ans[N],cnt[N];
13 int BLOCK;
14 bool cmp(node x,node y){
15     if(x.l/BLOCK==y.l/BLOCK) return x.r<y.r;
16     return x.l/BLOCKBLOCK;
17 }
18 
19 int n,m;
20 LL Ans=0;
21 
22 void add(int x){
23     Ans+=a[x]*(cnt[a[x]]*2+1);
24     cnt[a[x]]++;
25 }
26 
27 void del(int x){
28     Ans-=a[x]*(cnt[a[x]]*2-1);
29     cnt[a[x]]--;
30 }
31 
32 int main(){
33     scanf("%d%d",&n,&m);
34     BLOCK=sqrt(n);
35     for(int i=1;i<=n;i++){
36         scanf("%lld",&a[i]);
37     }
38     for(int i=1;i<=m;i++){
39         scanf("%d%d",&Q[i].l,&Q[i].r);
40         Q[i].id=i;
41     }
42     sort(Q+1,Q+1+m,cmp);
43     int L=1,R=0;
44     for(int i=1;i<=m;i++){
45         while(L<Q[i].l){
46             del(L);
47             L++;
48         }
49         while(L>Q[i].l){
50             L--;
51             add(L);
52         }
53         while(R<Q[i].r){
54             R++;
55             add(R);
56         }
57         while(R>Q[i].r){
58             del(R);
59             R--;
60         }
61         ans[Q[i].id]=Ans;
62     }
63     for(int i=1;i<=m;i++)
64     printf("%lld\n",ans[i]);
65     return 0;
66 }