莫队。
统计ai[i]的出现次数，每一次先还原贡献，再加上或减去当前的贡献即可。
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#include <iomanip>
#pragma GCC optimize(2)
#define up(i,a,b) for(int i=a;i<b;i++)
#define dw(i,a,b) for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
char ch = getchar(); ll x = 0, f = 1;
while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 2e5 + 10;
int ai[N], belong[N];
int n, t, block;
ll ans = 0;
int cnt[N * 10];
int L = 1, R;
struct node {
int l, r, id;
bool operator<(const node temp)const {
if (belong[l] == belong[temp.l])
{
if (belong[l] & 1)
{
return r < temp.r;
}
else return r > temp.r;
}
else return l < temp.l;
}
}qr[N];
void add(int x)
{
ans -= 1ll * cnt[x] * x* cnt[x];
cnt[x]++;
ans += 1ll * cnt[x] * x* cnt[x];
}
void del(int x)
{
ans -= 1ll * cnt[x] * x* cnt[x];
cnt[x]--;
ans += 1ll * cnt[x] * x* cnt[x];
}
ll ans_mo[N];
void mo()
{
upd(i, 1, t)
{
while (L < qr[i].l)del(ai[L++]);
while (L > qr[i].l)add(ai[--L]);
while (R < qr[i].r)add(ai[++R]);
while (R > qr[i].r)del(ai[R--]);
ans_mo[qr[i].id] = ans;
}
upd(i, 1, t)
{
cout << ans_mo[i] << endl;
}
}
int main()
{
cin >> n >> t;
block = sqrt(n);
upd(i, 1, n) {
cin >> ai[i];
belong[i] = (i - 1) / block + 1;
}
upd(i, 1, t)
{
qr[i].id = i;
qr[i].l = read();
qr[i].r = read();
}
sort(qr + 1, qr + 1 + t);
mo();
return 0;
}