codeforces 86 D. Powerful array(普通莫队)

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探讨了如何高效计算给定数组中多个子数组的能量，能量定义为子数组中每种数值出现频率的平方乘以其数值之和。采用预处理和优化搜索策略，避免重复计算，实现快速响应大量查询。

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D. Powerful array

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
3 2
1 2 1
1 2
1 3
output
3
6
input
8 3
1 1 2 2 1 3 1 1
2 7
1 6
2 7
output
20
20
20
Note
Consider the following array (see the second sample) and its [2, 7] subarray (elements of the subarray are colored):
在这里插入图片描述
Then K1 = 3, K2 = 2, K3 = 1, so the power is equal to 32·1 + 22·2 + 12·3 = 20.

**题解：题意就是给定一个区间然后对于区间里的每个数出现的次数，求sum[a[x]]sum[a[x]]a[x]
所以先将m个数排序，因为每次指针跳来跳去会浪费时间，这样可以减少次数，最后按序号再排序就行了，注意的是不能开太多long long 会T

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn =1e6 + 5;
int a[maxn];
ll ans;
ll sum[maxn];
int Be[maxn];
struct mo{
	int l, r;//左右区间
	int id;//序号
	ll s;//答案
}q[maxn];
bool cmp(mo a, mo b){
	if (Be[a.l] == Be[b.l])
		return a.r < b.r;
	return a.l < b.l;
}
bool CMP(mo a,mo b){
	return a.id < b.id;
}
void solve(int x, int add){
	ans -= sum[a[x]] * sum[a[x]]*a[x];
	sum[a[x]] += add;
	ans += sum[a[x]] * sum[a[x]]*a[x];
}
int main(){
	int n, m;
	cin >> n >> m;
	int unit = sqrt(n);
	for (int i = 1; i <= n; i++)
		scanf("%d",&a[i]), Be[i] = i / unit + 1;
	int l = 1, r = 0;
	for (int i = 1; i <= m; i++){
		scanf("%d%d", &q[i].l, &q[i].r);
		q[i].id = i;
	}
	sort(q + 1, q + 1 + m, cmp);
	for (int i = 1; i <= m; i++){
		while (l < q[i].l) solve(l, -1), l++;
		while (l>q[i].l) solve(l - 1, 1), l--;
		while (r < q[i].r) solve(r + 1, 1),r++;
		while (r>q[i].r) solve(r, -1), r--;
		q[i].s = ans;
	}
	sort(q + 1, q + 1 + m, CMP);
	for (int i = 1; i <= m; i++)
		printf("%I64d\n", q[i].s);
	return 0;
}