codeforces 86D D. Powerful array

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Kirito_Acmer

最新推荐文章于 2023-03-29 23:41:56 发布

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分类专栏： codeforces 莫队算法文章标签：莫队算法 codeforces

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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
这题也是用莫队算法，类型和前面小Z的袜子基本一样。
#include <cstdio>
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;

struct Query
{
    int id, l, r;
    long long ans;
};

const int MAXN = 200010;
const int MAXNUM = 1000010;
int n, m, sqrtn;
int c[MAXN], num[MAXNUM];
Query q[MAXN];

int gcd(int a, int b)
{
    return b == 0 ? a : gcd(b, a % b);
}

bool cmplr(const Query &a, const Query &b)
{
    if (a.l / sqrtn == b.l / sqrtn) return a.r < b.r;
    else return a.l < b.l;
}

bool cmpid(const Query &a, const Query &b)
{
    return a.id < b.id;
}

int main()
{
    scanf("%d%d", &n, &m);
    sqrtn = (int)sqrt(n);
    memset(num, 0, sizeof(num));
    for (int i = 1; i <= n; i++)
        scanf("%d", &c[i]);
    for (int i = 0; i < m; i++)
    {
        q[i].id = i;
        scanf("%d%d", &q[i].l, &q[i].r);
    }
    sort(q, q + m, cmplr);
    int l = 1, r = 1;
    long long ans = c[1];
    num[c[1]]++;
    for (int i = 0; i < m; i++)
    {
        while (r < q[i].r)
        {
            r++;
            ans -= (long long)num[c[r]] * num[c[r]] * c[r];
            num[c[r]]++;
            ans += (long long)num[c[r]] * num[c[r]] * c[r];
        }
        while (l < q[i].l)
        {
            ans -= (long long)num[c[l]] * num[c[l]] * c[l];
            num[c[l]]--;
            ans += (long long)num[c[l]] * num[c[l]] * c[l];
            l++;
        }
        while (l > q[i].l)
        {
            l--;
            ans -= (long long)num[c[l]] * num[c[l]] * c[l];
            num[c[l]]++;
            ans += (long long)num[c[l]] * num[c[l]] * c[l];
        }
        while (r > q[i].r)
        {
            ans -= (long long)num[c[r]] * num[c[r]] * c[r];
            num[c[r]]--;
            ans += (long long)num[c[r]] * num[c[r]] * c[r];
            r--;
        }
        q[i].ans = ans;
    }
    sort(q, q + m, cmpid);
    for (int i = 0; i < m; i++)
        cout << q[i].ans << "\n";
    return 0;
}
	 	 