莫队算法

简单版本：将区间分成sqrt(n)块并将询问按照左端点的所在块序号排序，左端点块序号一样则按照右端点排序

LL Gcd( LL a , LL b ){ return b==0?a:Gcd( b , a%b ); }
struct Query
{
    int L,R,id;
}Q[maxm];
struct Ans
{
    LL a,b;
    void reduce()
    {
        LL d = Gcd( a , b );
        a /= d;
        b /= d;
    }
}A[maxm];
int a[maxn],num[maxn],n,m,block;
bool cmp( Query a , Query b )
{
    if ( a.L/block!=b.L/block ) return a.L/block<b.L/block;
    else return a.R<b.R;
}
void work()
{
    LL tmp = 0; memset ( num , 0 , sizeof(num) );
    int L = 1,R = 0;
    for ( int i=0 ; i<m ; i++ )
    {
        while ( R<Q[i].R )
        {
            R++;
            tmp -= 1LL*num[a[R]]*num[a[R]];
            num[a[R]]++;
            tmp += 1LL*num[a[R]]*num[a[R]];
        }
        while ( R>Q[i].R )
        {
            tmp -= 1LL*num[a[R]]*num[a[R]];
            num[a[R]]--;
            tmp += 1LL*num[a[R]]*num[a[R]];
            R--;
        }
        while ( L<Q[i].L )
        {
            tmp -= 1LL*num[a[L]]*num[a[L]];
            num[a[L]]--;
            tmp += 1LL*num[a[L]]*num[a[L]];
            L++;
        }
        while ( L>Q[i].L )
        {
            L--;
            tmp -= 1LL*num[a[L]]*num[a[L]];
            num[a[L]]++;
            tmp += 1LL*num[a[L]]*num[a[L]];
        }
        A[Q[i].id].a = tmp-(R-L+1);
        A[Q[i].id].b = 1LL*(R-L+1)*(R-L);
        A[Q[i].id].reduce();
    }
}
int main()
{
    for ( ; scanf ( "%d%d" , &n , &m )==2 ; )
    {
        for ( int i=1 ; i<=n ; i++ )
            scanf ( "%d" , &a[i] );
        for ( int i=0 ; i< m ; i++ )
        {
            Q[i].id = i;
            scanf ( "%d%d" , &Q[i].L , &Q[i].R );
        }
        block = sqrt(n); sort ( Q , Q+m , cmp ); work();
        for ( int i=0 ; i<m ; i++ ) printf ( "%lld/%lld\n" , A[i].a , A[i].b );
    }
    return 0;
}