【P2709 小B的询问】莫队

P2709
不修莫队算法 维护一下就行了
这题让自己知道 初始坐标l r的设定和你扩张的顺序有很大关系

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define lc (rt<<1)
#define rc (rt<<11)
#define mid ((l+r)>>1)

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
const int MAX_N = 50025;
int cnt[MAX_N],arr[MAX_N],block;
ll Ans = 0,ans[MAX_N];
struct node
{
    int l,r,id;
}Q[MAX_N];
bool cmp(node a,node b)
{
    return (a.l/block)^(b.l/block)?a.l<b.l:(((a.l/block)&1)?a.r<b.r:a.r>b.r);
}
void add(int x)
{
    Ans-=cnt[arr[x]]*cnt[arr[x]];
    cnt[arr[x]]++;
    Ans+=cnt[arr[x]]*cnt[arr[x]];
}
void del(int x)
{
    Ans-=cnt[arr[x]]*cnt[arr[x]];
    cnt[arr[x]]--;
    Ans+=cnt[arr[x]]*cnt[arr[x]];
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,k,m;
    scanf("%d%d%d",&n,&m,&k);
    for(int i = 1;i<=n;++i) scanf("%d",&arr[i]);
    block = n/sqrt(m*2/3);
    for(int i = 1;i<=m;++i) scanf("%d%d",&Q[i].l,&Q[i].r),Q[i].id = i;
    sort(Q+1,Q+1+m,cmp);
    Ans= 0;int ql = 1,qr = 0;
    for(int i = 1;i<=m;++i)
    {
        while(qr<Q[i].r) add(++qr);
        while(ql>Q[i].l) add(--ql);
        while(qr>Q[i].r) del(qr--);
        while(ql<Q[i].l) del(ql++);
        ans[Q[i].id] = Ans;
    }
    for(int i = 1;i<=m;++i)
        printf("%lld\n",ans[i]);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}