给n个数， m个询问， 每次询问输出区间内的数的和， 相同的数只计算一次。
数组里的数是>-1e9 <1e9， 可以把它离散以后用莫队搞...
#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-;
const int mod = 1e9+;
const int inf = ;
const int dir[][] = { {-, }, {, }, {, -}, {, } };
const int maxn = 3e5+;
int a[maxn], b[maxn], c[maxn];
ll ans[];
struct node
{
int l, r, id, block;
bool operator < (node a)const
{
if(block == a.block)
return r<a.r;
return block<a.block;
}
}q[];
int main()
{
int t, n, m;
cin>>t;
while(t--) {
scanf("%d", &n);
mem(c);
for(int i = ; i<=n; i++) {
scanf("%d", &a[i]);
b[i-] = a[i];
}
sort(b, b+n);
int cnt = unique(b, b+n)-b;
for(int i = ; i<=n; i++) {
a[i] = lower_bound(b, b+cnt, a[i])-b+;
}
int BLOCK = max(1.0, sqrt(cnt*1.0));
cin>>m;
for(int i = ; i<m; i++) {
scanf("%d%d", &q[i].l, &q[i].r);
q[i].block = q[i].l/BLOCK;
q[i].id = i;
}
sort(q, q+m);
ll sum = ;
for(int i = q[].l; i<=q[].r; i++) {
if(!c[a[i]]) {
sum += b[a[i]-];
}
c[a[i]]++;
}
ans[q[].id] = sum;
for(int i = ; i<m; i++) {
if(q[i].l<q[i-].l) {
for(int j = q[i-].l-; j>=q[i].l; j--) {
if(!c[a[j]]) {
sum += b[a[j]-];
}
c[a[j]]++;
}
} else {
for(int j = q[i-].l; j<q[i].l; j++) {
if(c[a[j]]==) {
sum -= b[a[j]-];
}
c[a[j]]--;
}
}
if(q[i].r>q[i-].r) {
for(int j = q[i-].r+; j<=q[i].r; j++) {
if(!c[a[j]]) {
sum += b[a[j]-];
}
c[a[j]]++;
}
} else {
for(int j = q[i-].r; j>q[i].r; j--) {
if(c[a[j]] == ) {
sum -= b[a[j]-];
}
c[a[j]]--;
}
}
ans[q[i].id] = sum;
}
for(int i = ; i<m; i++)
printf("%I64d\n", ans[i]);
}
return ;
}