題目鏈接：http://codeforces.com/problemset/problem/86/D

題目：

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

You should calculate the power of t

First line contains two integers n and t (1 ≤ n, t ≤ 200000) — the array length and the number of queries correspondingly.

Second line contains n positive integers ai (1 ≤ ai ≤ 106) — the elements of the array.

Next t lines contain two positive integers l, r (1 ≤ l ≤ r ≤ n) each — the indices of the left and the right ends of the corresponding subarray.

Output t lines, the i-th line of the output should contain single positive integer — the power of the i-th query subarray.

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preferred to use cout stream (also you may use %I64d).

3 2
1 2 1
1 2
1 3

3
6

8 3
1 1 2 2 1 3 1 1
2 7
1 6
2 7

20
20
20

Consider the following array (see the second sample) and its [2, 7] subarray (elements of the subarray are colored):

題意：給出一個由n個正整數形成的數組，t次詢問，每次詢問一個區間［l，r］內所有 K^2＊a的和，K為數a在區間內出現的次數。

題解：更改下add和del即可。add的話（k+1）^2-k^2=2k+1，增加2k+1個a[i]；del的話k^2-(k-1)^2=2k-1，減少2k-1個a[i]。

 1 #include <cmath>
 2 #include <cstdio>
 3 #include <algorithm>
 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/BLOCK<y.l/BLOCK;
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 }

Codeforces 86D Powerful array（莫隊）