[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=,M=,B=;
 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,,M) fa[i]=i+; fa[M+]=M+;
     rep(i,,B) a[i]=i+;
     rep(i,,n){
         scanf(" %c",&ch);
         if (ch=='A'){
             scanf("%d",&x); fa[x]=x;
             rep(j,,B) a[j]=min(a[j],x%j);
             q[i]=; q2[i]=x;
         }else{
             scanf("%d",&x);
             if (x<=B) q[i]=,q2[i]=,ans[i]=a[x]; else q[i]=,q2[i]=x;
         }
     }
     for (int i=n; i; i--){
         if (q[i]) fa[q2[i]]=q2[i]+;
         else if (q2[i]){
             ans[i]=M+;
             for (int j=; j<=M; j+=q2[i])
                 if (get(j)<=M) ans[i]=min(ans[i],get(j)-j);
         }
     }
     rep(i,,n) if (!q[i]) printf("%d\n",ans[i]);
     return ;
 }