bzoj2038: [2009国家集训队]小Z的袜子(hose) 莫队

维护一个sum即分子,和一个计数用的数组即可O(1)转移

/**************************************************************
    Problem: 2038
    User: walfy
    Language: C++
    Result: Accepted
    Time:1896 ms
    Memory:3452 kb
****************************************************************/
 
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define fio ios::sync_with_stdio(false);cin.tie(0)
 
using namespace std;
 
const double eps=1e-6;
const int N=50000+10,maxn=20000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f;
 
int a[N],belong[N];
struct query{
    int l,r,id;
    bool operator <(const query&rhs)const{
        if(belong[l]==belong[rhs.l])return r<rhs.r;
        return l<rhs.l;
    }
}q[N];
ll sum,co[N];
pair<ll,ll>ans[N];
void gao(int pos,int op)
{
    if(op==1)
    {
        if(co[a[pos]]==0)co[a[pos]]++;
        else
        {
            sum-=co[a[pos]]*(co[a[pos]]-1);
            co[a[pos]]++;
            sum+=co[a[pos]]*(co[a[pos]]-1);
        }
    }
    else
    {
        if(co[a[pos]]==1)co[a[pos]]--;
        else
        {
            sum-=co[a[pos]]*(co[a[pos]]-1);
            co[a[pos]]--;
            sum+=co[a[pos]]*(co[a[pos]]-1);
        }
    }
}
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    int block=sqrt(n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        belong[i]=(i-1)/block+1;
    }
    for(int i=0;i<m;i++)
    {
        scanf("%d%d",&q[i].l,&q[i].r);
        q[i].id=i;
    }
    sort(q,q+m);
    int l=1,r=1;co[a[1]]=1;
    for(int i=0;i<m;i++)
    {
        while(l>q[i].l)l--,gao(l,1);
        while(r<q[i].r)r++,gao(r,1);
        while(l<q[i].l)gao(l,-1),l++;
        while(r>q[i].r)gao(r,-1),r--;
//        printf("%d -----%d %lld
",l,r,sum);
//        for(int j=0;j<=n;j++)printf("%d ",co[j]);puts("");
        ll fz=sum,fm=1ll*(r-l+1)*(r-l),te=__gcd(fz,fm);
        if(fm==0)
        {
            ans[q[i].id]=mp(0,1);
            continue;
        }
//        printf("%lld %lld %lld
",fz,fm,te);
        fz/=te,fm/=te;
        ans[q[i].id]=mp(fz,fm);
    }
    for(int i=0;i<m;i++)printf("%lld/%lld
",ans[i].fi,ans[i].se);
    return 0;
}
/********************
 
********************/