POJ2886（反素数+约瑟夫环+线段树）#nobody

反素数f(x)：是指不大与x约数最多的数

反素数的程序见我这篇博客：http://blog.csdn.net/xdu_truth/article/details/8043051

先打表出来反素数，那么我们就可以知道n个人第几个人出来事最大的。

然后就是个约瑟夫环问题，但是这题数据范围较大，所以就要用到线段树查找删除。

CODE：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

int p[35] = {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};
int np[35] = {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};
char s[500010][15];
int num[500010];
int ans,n,k;
int find(int n)
{
    int l = 0,r = 34;
    while(l<=r)
    {
        int m = r+l>>1;
        if(p[m]>n) r = m-1;
        else l = m+1;
    }
    return l-1;
}
int SUM[500001<<2];
void pushup(int rt)
{
    SUM[rt] = SUM[rt<<1] + SUM[rt<<1|1];
}
void build(int l,int r,int rt)
{
    if(l == r && l == k){SUM[rt] = 0;return ;}
    if(l == r){SUM[rt] = 1;return;}
    int m = r+l >> 1;
    build(lson);
    build(rson);
    pushup(rt);
}
void update(int p,int l,int r,int rt)
{
    if(l == r){SUM[rt] = 0;ans = l;return;}
    int m = l+r>>1;
    if(SUM[rt<<1] >= p){update(p,lson);pushup(rt);}
    else
    {
        p -= SUM[rt<<1];
        update(p,rson);
        pushup(rt);
    }
}
int main()
{
    //freopen("input.txt","r",stdin);
    while(~scanf("%d%d",&n,&k))
    {
        int r = find(n);
        for(int i=1;i<=n;i++)
            scanf("%s%d",s[i],&num[i]);
        int now = k;
        build(1,n,1);
        int nn = n-1;
        ans = k;
        for(int i=1;i<p[r];i++)
        {
            int t = num[ans] > 0 ? num[ans]+now-1:num[ans]+now;
            t--;
            while(t<0)t+=nn;
            t = t%nn;
            t++;
            //cout << t<< " "<< endl;
            now = t;
            update(t,1,n,1);
            nn--;
        }
        printf("%s %d\n",s[ans],np[r]);
    }
    return 0;
}