思路

很显然一眼看过去就是一道很经典的二维偏序问题，我们通过对询问存储下来然后升序排序，通过线段树或者树状数组更新并离线询问即可

Code

#include<bits/stdc++.h>
#include<ctime>
#include<cstdlib>
using namespace std;
#define ll long long
const int N = 5e5+10;
ll tree[N],a[N],ans[N],vis[N];
ll n,m;

struct Node {
	int pos,l,r;
}V[N];

ll lowbit(ll x) {
	return -x & x;
} 

void updata(ll x,ll k) {
	while(x <= n) {
		tree[x] += k;
		x += lowbit(x);
	}
}
ll query(ll x){
	ll ans = 0;
	while(x) {
		ans += tree[x];
		x -= lowbit(x);
	}
	return ans;
}

int main()
{

	scanf("%lld%lld",&n,&m);
	for(int i = 1;i <= n; ++i) {
		scanf("%lld",&a[i]);
	}
	ll l,r;
	for(ll i = 1;i <= m; ++i) {
		scanf("%lld%lld",&l,&r);
		V[i].l = l;
		V[i].r = r;
		V[i].pos = i;
	}
	sort(V+1,V+1+m,[](Node a,Node b){return a.r == b.r?a.l < b.l:a.r < b.r;});
	ll loc = 1;
	for(ll i = 1;i <= m; ++i) {
		for(ll j = loc;j <= V[i].r; ++j) {
			if(vis[a[j]]) updata(vis[a[j]],-a[j]);
			vis[a[j]] = j;
			updata(j,a[j]);
		}
		loc = V[i].r + 1;
		ans[V[i].pos] = query(V[i].r) - query(V[i].l - 1);
	}
	for(ll i = 1;i <= m; ++i) {
		printf("%lld\n",ans[i]);
	}
	return 0;
 }