給你n個數，m次詢問，Ks為區間內s的數目，求區間[L,R]之間所有Ks*Ks*s的和。1<=n，m<=200000.1<=s<=10^6

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include 
 
<cmath>
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 200050, INF = 0x7fffffff;
int n, m, pos[maxn], s[maxn*10
 
], c[maxn];
LL ans;

struct node
{
    int l, r, id;
    LL res;
}Node[maxn];

bool cmp(node a, node b)
{
    return pos[a.l] == pos[b.l] ? (a.r < b.r) : (a.l < b.l);
}

bool cmp_id(node a, node b)
{
    return a.id < b.id;
}



void add(LL x)
{
    s[c[x]]++;
    ans += c[x]*(s[c[x]]*s[c[x]] - (s[c[x]]-1
 
)*(s[c[x]]-1));
}

void del(LL x)
{
    s[c[x]]--;
    ans -= c[x]*((s[c[x]]+1)*(s[c[x]]+1) - (s[c[x]])*(s[c[x]]));
}


int main()
{
    ans = 0;
    scanf("%d%d", &n, &m);
    for(int i=1; i<=n; i++)
        scanf("%d", &c[i]);
    int block = sqrt(n);
    for(int i=1; i<=n; i++)
        pos[i] = (i-1)/block + 1;
    for(int i=1; i<=m; i++)
    {
        scanf("%d%d", &Node[i].l, &Node[i].r);
        Node[i].id = i;
    }
    sort(Node+1, Node+1+m, cmp);
    for(int i=1, l=1, r=0; i<=m; i++)
    {
        for(; r < Node[i].r; r++)
            add(r+1);
        for(; r > Node[i].r; r--)
            del(r);
        for(; l < Node[i].l; l++)
            del(l);
        for(; l > Node[i].l; l--)
            add(l-1);
        Node[i].res = ans;
    }
    sort(Node+1, Node+1+m, cmp_id);
    for(int i=1; i<=m; i++)
        printf("%I64d\n", Node[i].res);


    return 0;
}

Powerful array CodeForces - 86D（莫隊）