线段树点更新+反素数+poj2777

来源：互联网发布：国内外大数据公司编辑：程序博客网时间：2024/05/17 03:26

反素数：对于任何正整数x,其约数的个数记做g(x).例如g(1)=1,g(6)=4.如果某个正整数x满足：对于任意i(0<i<x),都有g(i)<g(x),则称x为反素数。

思路：打表，线段树维护区间空位数。

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<cmath>#include<string>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;const int antiprime[36] = {1,2,4,6,12,24,36,48,60,120,180,240,360,720,840,1260,1680,2520,5040,7560,10080,15120,20160,25200,27720,45360,50400,55440,83160,110880,166320,221760,277200,332640,498960,554400};const int factor[36] = {1,2,3,4,6,8,9,10,12,16,18,20,24,30,32,36,40,48,60,64,72,80,84,90,96,100,108,120,128,144,160,168,180,192,200,216};const int maxn=500010;char name[maxn][15];int a[maxn];int N,K;struct IntervalTree{    int sum[maxn<<3];    void build(int o,int l,int r)    {        sum[o]=r-l+1;        if(l==r)return;        int mid=(l+r)>>1;        build(o<<1,l,mid);        build(o<<1|1,mid+1,r);    }    void update(int o,int l,int r,int pos,int x)    {        if(l==r)        {            sum[o]+=x;            return ;        }        int mid=(l+r)>>1;        if(pos<=mid)update(o<<1,l,mid,pos,x);        else update(o<<1|1,mid+1,r,pos,x);        sum[o]=sum[o<<1]+sum[o<<1|1];    }    int query(int o,int l,int r,int x)    {        if(sum[o]==x&&l==r)return l;        int mid=(l+r)>>1;        if(sum[o<<1]>=x)return query(o<<1,l,mid,x);        return query(o<<1|1,mid+1,r,x-sum[o<<1]);    }}tree;int main(){    //freopen("in.txt","r",stdin);    while(scanf("%d%d",&N,&K)!=EOF)    {        for(int i=1;i<=N;i++)            scanf("%s%d",name[i],&a[i]);        tree.build(1,1,N);        int step=0,pos;        for(int i=0;i<36;i++)            if(antiprime[i]<=N)            {                step=antiprime[i];                pos=i;            }        int cur=K,x=K;        tree.update(1,1,N,K,-1);        for(int i=1;i<step;i++)        {            int ans;            if(a[x]>0)ans=(cur+a[x]-2)%(N-i)+1;            else                ans=((cur+a[x]-1)%(N-i)+(N-i))%(N-i)+1;            if(!ans)ans=1;            x=tree.query(1,1,N,ans);            tree.update(1,1,N,x,-1);            cur=ans;        }        printf("%s %d\n",name[x],factor[pos]);    }    return 0;}

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