莫队算法-Powerful array

Powerful array

time limit per test

5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

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

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).

Examples

input

Copy

3 2
1 2 1
1 2
1 3

output

Copy

3
6

input

Copy

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

output

Copy

20
20
20

Note

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

Then K1 = 3, K2 = 2, K3 = 1, so the power is equal to 32·1 + 22·2 + 12·3 = 20.

 题意大概是这样的，给出一个数组，对每次询问的区间进行计算，公式为数字出现次数的平方乘数字。

题意很好理解，主要问题是数据规模太大，数组大小和询问次数都是2e5。第一反应是前缀和，然后mle，试了一下分块，还是mle。补题的时候才看到莫队算法，学到了。

算法的大意是，对于密集的区间查询，可以尽可能利用已有的查询结果，修修补补凑合着用。

代码如下，内容参考https://blog.csdn.net/vaeloverforever/article/details/82314626

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 50;
const int INF = 0xffffff;
 
int n,t;
int arr[maxn];
int cnt[1000005], id[maxn];
ll ans[maxn];
int L,R;
int unit;
ll now;
 
struct range{
    int l,r;
    int askorder;
    friend bool operator < (const range &a, const range &b){
        return a.l/unit != b.l/unit ? a.l/unit < b.l/unit : a.r < b.r;    
    }
}rangee[maxn];
 
void del(int x){//删掉不在区间里的
    now -= ((cnt[arr[x]]<<1)-1)*(ll)arr[x];
        cnt[arr[x]]--;
} 
void add(int x){//加上进入区间的
    now += ((cnt[arr[x]]<<1)|1)*(ll)arr[x];
    cnt[arr[x]]++;
}
 
int main()
{
    scanf("%d%d",&n,&t);
    for(int i = 1;i < n+1; ++i)
        scanf("%d",&arr[i]);
    unit = sqrt(t*2.0/3);//块大小
    for(int i = 0; i < t; ++i){
        scanf("%d%d",&rangee[i].l,&rangee[i].r);
        rangee[i].askorder = i;
    }
    sort(rangee, rangee+t);//排序以尽可能复用先前结果
    for(int i = 0; i < t; ++i){
        while (L < rangee[i].l) 
            del(L++);
        while (R > rangee[i].r) 
            del(R--);
        
        while (L > rangee[i].l) 
            add(--L);
        while (R < rangee[i].r) 
            add(++R);
        ans[rangee[i].askorder] = now;
    }
    for(int i = 0;i < t; ++i)
        printf("%lld\n",ans[i]);
    return 0;
}