小Z的袜子

Posted 2022-02-20 rign

将询问排序分块后每块内部排序

然后每次左端点增减复杂度为√N，每块右端点增减N

每次问询复杂度√N+N

总共有√N次问询

#include <bits/stdc++.h>
using namespace std;
#define fore(i,a,b) for(int i=(int)(a);i<=(int)(b);i++)
#define forb(i,a,b) for(int i=(int)(a);i>=(int)(b);i--)
#define fir first
#define sec second
typedef long long ll;

const int N=500010;
int c[N],L[N],R[N],num[N];
struct nodeint l,r,id;q[N];
bool cmp(node a,node b)return a.l<b.l || (a.l==b.l && a.r<b.r);
bool cmp0(node a,node b)return a.r<b.r;
ll ans,anss[N][2],g;
int T, n, m;

void work(int x,int w)ans-=(ll)num[x]*(num[x]-1);num[x]+=w;ans+=(ll)num[x]*(num[x]-1);
ll gcd(ll a,ll b)return b?gcd(b,a%b):a;
void gcdab(ll &a,ll &b)g=gcd(a,b);if(!g) b=1; else a/=g,b/=g;

int main()
    cin>>n>>m;
    fore(i,1,n)
        scanf("%d",&c[i]);
    fore(i,1,m)
        scanf("%d%d",&q[i].l,&q[i].r);
        q[i].id=i;
    
    sort(q+1,q+m+1,cmp);
    int t=sqrt(m);
    fore(i,1,t)
        L[i]=(i-1)*t+1,R[i]=i*t;
    if(R[t]<n) L[t+1]=R[t]+1,R[++t]=m;
    fore(i,1,t)
        memset(num,0,sizeof(num));
        sort(q+L[i],q+R[i]+1,cmp0);
        ans=0;
        int l=q[L[i]].l,r=q[L[i]].r;
        fore(j,l,r)
            work(c[j],1);
        anss[q[L[i]].id][0]=ans;
        anss[q[L[i]].id][1]=(ll)(r-l)*(r-l+1);
        gcdab(anss[q[L[i]].id][0],anss[q[L[i]].id][1]);
        fore(j,L[i]+1,R[i])
            while(r<q[j].r) work(c[++r],1);
            while(r>q[j].r) work(c[r--],-1);
            while(l<q[j].l) work(c[l++],-1);
            while(l>q[j].l) work(c[--l],1);
            if(q[j].l==q[j].r) anss[q[j].id][0]=0,anss[q[j].id][1]=1;
            else
                anss[q[j].id][0]=ans;
                anss[q[j].id][1]=(ll)(r-l)*(r-l+1);
                gcdab(anss[q[j].id][0],anss[q[j].id][1]);
            
        
    
    fore(i,1,m)
        printf("%lld/%lld\n",anss[i][0],anss[i][1]);
    return 0;