题目传送门：https://codeforces.com/problemset/problem/86/D
题目大意：
给定一个长度为(n)的序列，有(m)组询问，每次询问([l,r])中，(sumlimits_{s}K_s^2 imes s)的值，其中，(K_s)表示(s)在子串([l,r])中的出现次数
由于这题没有修改，仅有询问，故我们可以考虑莫队算法
/*program from Wolfycz*/
#include<map>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define Fi first
#define Se second
#define ll_inf 1e18
#define MK make_pair
#define sqr(x) ((x)*(x))
#define pii pair<int,int>
#define int_inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
static char buf[1000000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
template<typename T>inline T frd(T x){
int f=1; char ch=gc();
for (;ch<'0'||ch>'9';ch=gc()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=gc()) x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
template<typename T>inline T read(T x){
int f=1; char ch=getchar();
for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
inline void print(int x){
if (x<0) putchar('-'),x=-x;
if (x>9) print(x/10);
putchar(x%10+'0');
}
const int N=2e5,M=1e6;
int A[N+10],pos[N+10];
struct node{
int l,r,ID;
void Read(int i){l=read(0),r=read(0),ID=i;}
node(int _l=0,int _r=0,int _ID=0){l=_l,r=_r,ID=_ID;}
bool operator <(const node &tis)const{return pos[l]!=pos[tis.l]?l<tis.l:r<tis.r;}
}B[N+10];
ll All,Ans[N+10],Cnt[M+10];
void Add(int x,int v){
All-=1ll*x*sqr(Cnt[x]);
Cnt[x]+=v;
All+=1ll*x*sqr(Cnt[x]);
}
int main(){
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
int n=read(0),m=read(0),size=sqrt(n);
for (int i=1;i<=n;i++) A[i]=read(0),pos[i]=(i-1)/size+1;
for (int i=1;i<=m;i++) B[i].Read(i);
sort(B+1,B+1+m);
for (int i=1,l=1,r=0;i<=m;i++){
while (r<B[i].r) Add(A[++r], 1);
while (r>B[i].r) Add(A[r--],-1);
while (l<B[i].l) Add(A[l++],-1);
while (l>B[i].l) Add(A[--l], 1);
Ans[B[i].ID]=All;
// printf("
");
}
for (int i=1;i<=m;i++) printf("%lld
",Ans[i]);
return 0;
}