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直接模拟,把平衡树当做可以\Theta(\log n)访问并删除任意位置的链表

#include<bits/stdc++.h>
#pragma GCC optimize(3)
#define il __inline__ __attribute__ ((always_inline))
using namespace std;
namespace IO{const int str=1<<20;static char in_buf[str],*in_s,*in_t;bool __=0;il char gc(){return (in_s==in_t)&&(in_t=(in_s=in_buf)+fread(in_buf,1,str,stdin)),in_s==in_t?__=1,EOF:*in_s++;}il void in(string &ch){ch.clear();if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)__=1;}il void in(char &ch){if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)__=1;else ch=c;}il void in(char *ch){*ch='\0';if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)__=1;*ch='\0';}template<typename T>il void in(T &x){if(__)return;char c=gc();bool f=0;while(c!=EOF&&(c<'0'||c>'9'))f^=(c=='-'),c=gc();if(c==EOF){__=1;return;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=x*10+c-48,c=gc();if(c==EOF)__=1;if(f)x=-x;}template<typename T,typename ... arr>il void in(T &x,arr & ... y){in(x),in(y...);}const char ln='\n';static char out_buf[str],*out_s=out_buf,*out_t=out_buf+str;il void flush(){fwrite(out_buf,1,out_s-out_buf,stdout);out_s=out_buf;}il void pt(char c){(out_s==out_t)?(fwrite(out_s=out_buf,1,str,stdout),*out_s++=c):(*out_s++=c);}il void out(const char* s){while(*s)pt(*s++);}il void out(char* s){while(*s)pt(*s++);}il void out(char c){pt(c);}il void out(string s){for(int i=0;s[i];i++)pt(s[i]);}template<typename T>il void out(T x){if(!x){pt('0');return;}if(x<0)pt('-'),x=-x;char a[50],t=0;while(x)a[t++]=x%10,x/= 10;while(t--)pt(a[t]+'0');}template<typename T,typename ... arr>il void out(T x,arr & ... y){out(x),out(y...);}}using namespace IO;
const int N=700011;
#define ls c[rt][0]
#define rs c[rt][1]
int n,q,sz[N],c[N][2],val[N],rnd[N],rev[N],RT=0,cnt=0;
il int ne(int v){
    val[++cnt]=v;
    sz[cnt]=1;
    rnd[cnt]=rand()<<15|rand();
    return cnt;
}
il void pu(int rt){sz[rt]=sz[ls]+sz[rs]+1;}
void sl(int rt,int k,int &x,int &y){
    if(!rt)return (void)(x=y=0);
    if(sz[ls]>=k)y=rt,sl(ls,k,x,ls);
    else x=rt,sl(rs,k-sz[ls]-1,rs,y);
    pu(rt);
}
int mg(int x,int y){
    if(!x||!y)return x+y;
    if(rnd[x]<rnd[y]){
        c[x][1]=mg(c[x][1],y);
        pu(x);return x;
    }
    else{
        c[y][0]=mg(x,c[y][0]);
        pu(y);return y;
    }
}
il int get(int p){
    int x,y,z,t;
    sl(RT,p,x,z);
    sl(x,p-1,x,y);
    t=val[y];
    RT=mg(mg(x,mg(c[y][0],c[y][1])),z);
    return t;
}
int main(){
    srand(time(0));
    in(n);
    int x,cur=1;
    for(int i=1;i<=n;++i)RT=mg(RT,ne(i));
    for(int i=1;i<=n;++i){
        in(x);
        cur=(cur+x-1)%(n-i+1)+1;
        out(get(cur),ln);
    }
    flush();
}
LG 3988 [SHOI2013]发牌
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