(中等) POJ 2886 Who Gets the Most Candies? , 反素数+线段树。

Description

　　N children are sitting in a circle to play a game.

　　The children are numbered from 1 to N in clockwise order. Each of them has a card with a non-zero integer on it in his/her hand. The game starts from the K-th child, who tells all the others the integer on his card and jumps out of the circle. The integer on his card tells the next child to jump out. Let A denote the integer. If A is positive, the next child will be the A-th child to the left. If A is negative, the next child will be the (−A)-th child to the right.

　　The game lasts until all children have jumped out of the circle. During the game, the p-th child jumping out will get F(p) candies where F(p) is the number of positive integers that perfectly divide p. Who gets the most candies?

 

　　题意大致等同于约瑟夫环的问题，一次出来一个，然后出来x个人，其中x表示约数个数最多的数中最小的那个。

 

　　做这个题时还不知道什么是反素数，用最笨的方法 (用了3s+时间) 打出了表，直接复制上的。然后就是构建线段树了，这个的线段树不难构建，就是记录区间内还没出去的人的个数。然后更新的话是标记某个点为出去了，并且返回这个点的位置，作为下一次计数的起点。询问的话就是某个区间的没出去的个数。

 

　　反素数的话也在学习中，推荐ACdreamer的文章： http://blog.csdn.net/ACdreamers/article/details/25049767

 

代码如下：（注：写的比较乱，水平有限。）

#include<iostream>
#include<cstdio>

using namespace std;

const int remMax[36][2]={
    {1,1},{2,2},{4,3},{6,4},{12,6},
    {24,8},{36,9},{48,10},{60,12},
    {120,16},{180,18},{240,20},{360,24},
    {720,30},{840,32},{1260,36},{1680,40},
    {2520,48},{5040,60},{7560,64},{10080,72},
    {15120,80},{20160,84},{25200,90},{27720,96},
    {45360,100},{50400,108},{55440,120},{83160,128},
    {110880,144},{166320,160},{221760,168},{277200,180},
    {332640,192},{498960,200},{500001,200}};

int N,K;
char name[500005][14];
int number[500005];
int BIT[500005*4];

int findM(int x)
{
    for(int i=0;i<35;++i)
        if(remMax[i][0]<=x&&remMax[i+1][0]>x)
            return i;
}

void pushUP(int po)
{
    BIT[po]=BIT[po*2]+BIT[po*2+1];
}

void build_tree(int L,int R,int po)
{
    BIT[po]=(R-L+1);

    if(R==L) return;

    int M=(L+R)/2;
    build_tree(L,M,po*2);
    build_tree(M+1,R,po*2+1);
}

int query(int ql,int qr,int L,int R,int po)
{
    if(ql>qr)
        return 0;

    if(ql<=L&&qr>=R)
        return BIT[po];

    int M=(L+R)/2;
    int temp=0;

    if(ql<=M)
        temp+=query(ql,qr,L,M,po*2);
    if(qr>M)
        temp+=query(ql,qr,M+1,R,po*2+1);

    return temp;
}

int update(int un,int L,int R,int po)
{
    --BIT[po];

    if(L==R)
        return L;

    int M=(L+R)/2;

    if(BIT[po*2]>=un)
        return update(un,L,M,po*2);
    else
        return update(un-BIT[po*2],M+1,R,po*2+1);
}

int main()
{
    int temp,temp1;
    int n,las;
    int times,fp;

    while(~scanf("%d %d",&N,&K))
    {
        for(int i=1;i<=N;++i)
            scanf("%s %d",name[i],&number[i]);

        build_tree(1,N,1);

        temp=findM(N);
        times=remMax[temp][0];
        fp=remMax[temp][1];

        las=0;

        for(int i=1;i<=times;++i)
        {
            if(K>0)
            {
                K%=(N-i+1);
                if(K==0)
                    K=N-i+1;
            }
            else
            {
                K%=(N-i+1);
                if(K==0)
                    K=1;
                else
                    K=(N-i+2)+K;
            }

            temp1=query(las+1,N,1,N,1);

            if(temp1>=K)
                K=(N-i+1)-(temp1-K);
            else
                K=(K-temp1);

            temp1=update(K,1,N,1);

            las=temp1;
            K=number[las];
        }

        printf("%s %d\n",name[las],fp);
    }

    return 0;
}