CF 86D Powerful array

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本文深入探讨了计算机科学中高效算法与数据结构的应用，包括但不限于排序算法、动态规划、哈希算法等核心概念。通过实际案例分析，展示了这些原理在解决复杂问题时的优势和效率提升。同时，文章还介绍了如何在编程实践中运用这些知识，为开发者提供实用指导。

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 productsKs·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).

Sample test(s)

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):

$\text{[math]}$ Then K1 = 3, K2 = 2, K3 = 1, so the power is equal to 32·1 + 22·2 + 12·3 = 20.

题意：有N个数， t个询问。每次输出区间的每个数字出现的次数^ 2 * 该数字的总和。

思路：莫队算法

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

// n * n * a - x = (n - 1) * (n - 1) * a; ->  x = (2 * n - 1) * a;
// n * n * a + x = (n + 1) * (n + 1) * a; ->  x = (2 * n + 1) * a;

const int V = 200000 + 50;
const int MaxN = 1000000 + 50;
const int mod = 1000000000 + 7;
struct node {
    int id, l, r;
};
node nod[V];
int n, t, num[V], L, R, sum[MaxN];
__int64 ans[V], now;
bool cmp(node a, node b) {
    int m = (int) sqrt(n);
    if(a.l / m != b.l / m)
        return a.l < b.l;
    return a.r < b.r;
}
void f(int l, int r) {
    while(L < l) {
        now -= num[L] * (2 * sum[num[L]]-- - 1);
        L++;
    }
    while(R > r) {
        now -= num[R] * (2 * sum[num[R]]-- - 1);
        R--;
    }
    while(L > l) {
        L--;
        now += num[L] * (2 * sum[num[L]]++ + 1);
    }
    while(R < r) {
        R++;
        now += num[R] * (2 * sum[num[R]]++ + 1);
    }
}
void Modui() {
    now = L = R = 0;
    for(int i = 0; i < t; ++i) {
        f(nod[i].l, nod[i].r);
        ans[nod[i].id] = now;
    }
}
int main() {
    int i, j;
    scanf("%d%d", &n, &t);
    for(i = 1; i <= n; ++i)
        scanf("%d", &num[i]);
    for(i = 0; i < t; ++i) {
        int l, r;
        nod[i].id = i;
        scanf("%d%d", &nod[i].l, &nod[i].r);
    }
    sort(nod, nod + t, cmp);
    Modui();
    for(i = 0; i < t; ++i)
        printf("%I64d\n", ans[i]);
}