poj2886

此題需要了解反素數的知識，之前不瞭解反素數還想把1e5以內的數的因子個數全部打表。。。。

此題還有一點，就是線段樹的更新操作，刪除某節點，這可能是大家之前沒有接觸到得

#include <iostream>
#include<stdio.h>
#include<algorithm>
#include<cstring>
using namespace std;
const int MAX=5*1e5+20;
typedef struct
{
    char name[15];
    int v;
}P;
P p[MAX];
int sum[4*MAX];
int N,K;
int RPrime[]={//反素數
    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
};
int fact[]={//反素數約數個數
    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
};
void build(int L,int R,int id)
{
    sum[id]=R-L+1;
    if(L==R)
        return;
    int mid=(L+R)>>1;
    build(L,mid,id*2);
    build(mid+1,R,id*2+1);
}
int update(int p,int L,int R,int id)
{
    sum[id]--;
    if(L==R)
        return L;
    int mid=(L+R)>>1;
    if(p<=sum[id*2])
        return update(p,L,mid,id*2);
    else
        return update(p-sum[id*2],mid+1,R,id*2+1);
}
int main()
{
    int i,j,cnt;
    while(~scanf("%d%d",&N,&K))
    {
        memset(sum,0,sizeof(sum));
        cnt=0;
        for(i=1;i<=N;i++)
            scanf("%s%d",p[i].name,&p[i].v);
        build(1,N,1);
        for(int i=0;i<=35;i++)
        {
            if(RPrime[i]<=N)
                cnt++;
        }
        cnt--;
        int na,h=N;
        for(i=1;i<=RPrime[cnt];i++)
        {
            na=update(K,1,N,1);
            if(--h>0)
            {
            int v=p[na].v;
            if(v>=0)
            {
                K=(K+v-1)%h;
                if(K==0)
                    K+=h;
            }
            else
            {
                K=(K+v)%h;
                if(K<=0)
                    K+=h;
            }
            }
        }
        cout<<p[na].name<<" "<<fact[cnt]<<endl;
    }
    return 0;
}