[bzoj] 2038 小Z的袜子(hose) || 莫队

#include<cstdio>
#include<algorithm>
#include<cmath>
#define N 50010
typedef long long ll;
using namespace std;
struct hhh
{
    int x,y,l,id;
    bool operator < (const hhh &b) const
    {
        if (b.l==l) return y<b.y;
        return l<b.l;
    }
}q[N];
ll n,m,s,l,r,a[N],ans1[N],ans2[N],cur,cnt[N];

ll read()
{
    ll ans=0,fu=1;
    char j=getchar();
    for (;j<'0' || j>'9';j=getchar()) if (j=='-') fu=-1;
    for (;j>='0' && j<='9';j=getchar()) ans*=10,ans+=j-'0';
    return ans*fu;
}

ll gcd(ll x,ll y)
{
    return y?gcd(y,x%y):x;
}

int main()
{
    n=read();s=sqrt(n);cnt[0]=1;
    m=read();
    for (int i=1;i<=n;i++) a[i]=read();
    for (int i=1;i<=m;i++)
    {
    q[i].x=read();
    q[i].y=read();
    q[i].id=i;
    q[i].l=(q[i].x-1)/s+1;
    }
    sort(q+1,q+m+1);
    for (int i=1;i<=m;i++)
    {
    while (l<q[i].x) cur-=--cnt[a[l++]];
    while (l>q[i].x) cur+=cnt[a[--l]]++;
    while (r<q[i].y) cur+=cnt[a[++r]]++;
    while (r>q[i].y) cur-=--cnt[a[r--]];
    ans1[q[i].id]=cur;
    ans2[q[i].id]=1LL*(r-l+1)*(r-l)/2;
    }
    for (int i=1;i<=m;i++)
    {
    ll g=gcd(ans1[i],ans2[i]);
    printf("%lld/%lld\n",ans1[i]/g,ans2[i]/g);
    }
    return 0;
}