[BZOJ4320][ShangHai2006]Homework(根号分治+并查集)

对于<=sqrt(300000)的询问，对每个模数直接记录结果，每次加入新数时暴力更新每个模数的结果。

对于>sqrt(300000)的询问，枚举倍数，每次查询大于等于这个倍数的最小数是多少，这个操作通过将询问逆序使用并查集支持。

#include<cstdio>
#include<algorithm>
#define rep(i,l,r) for (int i=(l); i<=(r); i++)
using namespace std;

const int N=300010,M=300000,B=550;
char ch;
int n,x,a[N],ans[N],q[N],q2[N],fa[N];
int get(int x){ return (fa[x]==x) ? x : fa[x]=get(fa[x]); }

int main(){
    freopen("bzoj4320.in","r",stdin);
    freopen("bzoj4320.out","w",stdout);
    scanf("%d",&n);
    rep(i,0,M) fa[i]=i+1; fa[M+1]=M+1;
    rep(i,1,B) a[i]=i+1;
    rep(i,1,n){
        scanf(" %c",&ch);
        if (ch=='A'){
            scanf("%d",&x); fa[x]=x;
            rep(j,1,B) a[j]=min(a[j],x%j);
            q[i]=1; q2[i]=x;
        }else{
            scanf("%d",&x);
            if (x<=B) q[i]=0,q2[i]=0,ans[i]=a[x]; else q[i]=0,q2[i]=x;
        }
    }
    for (int i=n; i; i--){
        if (q[i]) fa[q2[i]]=q2[i]+1;
        else if (q2[i]){
            ans[i]=M+1;
            for (int j=0; j<=M; j+=q2[i])
                if (get(j)<=M) ans[i]=min(ans[i],get(j)-j);
        }
    }
    rep(i,1,n) if (!q[i]) printf("%d
",ans[i]);
    return 0;
}