BZOJ4320 ShangHai2006 Homework

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考虑对询问大于根号300000和小于根号300000的分开做，对于小于根号300000的，直接开个数组记录答案，插入的时候O(根号300000)更新，对于大于的，假设询问数为k，我们枚举j=ik,i>=0,ik<=300000，然后找到j大于等于j的第一个数，更新答案

这个我们可以倒着做，每个点用并查集指向大于等于他的第一个数，倒着做把加入变成删除，删除的时候就直接把x这个点的根指向x+1的根

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<ctime>#include<cmath>#include<algorithm>#include<iomanip>#include<vector>#include<map>#include<set>#include<bitset>#include<queue>#include<stack>using namespace std;#define MAXN 100010#define MAXM 300010#define MAXS 1010#define INF 1000000000#define MOD 1000000007#define eps 1e-8#define ll long longint x[MAXN];char o[MAXN][2];bool v[MAXM];int f[MAXM];int s[MAXS];int n;int lim=300000;int ans[MAXN];int siz;int fa(int x){return f[x]==x?x:f[x]=fa(f[x]);}int main(){int i,j;scanf("%d",&n);memset(s,0x3f,sizeof(s));siz=sqrt(lim);for(i=1;i<=n;i++){scanf("%s%d",o[i],&x[i]);if(o[i][0]=='A'){v[x[i]]=1;for(j=1;j<=siz;j++){s[j]=min(s[j],x[i]%j);}}if(o[i][0]=='B'){if(x[i]<=siz){ans[i]=s[x[i]];}}}f[lim+1]=lim+1;for(i=lim;~i;i--){if(v[i]){f[i]=i;}else{f[i]=fa(i+1);}}for(i=n;i;i--){if(o[i][0]=='A'){f[fa(x[i])]=fa(x[i]+1);}if(o[i][0]=='B'){if(x[i]>siz){ans[i]=INF;for(j=0;j<=lim;j+=x[i]){if(fa(j)!=lim+1){ans[i]=min(ans[i],fa(j)-j);}}}}}for(i=1;i<=n;i++){if(o[i][0]=='B'){printf("%d\n",ans[i]);}}return 0;}/**/

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