(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[][]={
    {,},{,},{,},{,},{,},
    {,},{,},{,},{,},
    {,},{,},{,},{,},
    {,},{,},{,},{,},
    {,},{,},{,},{,},
    {,},{,},{,},{,},
    {,},{,},{,},{,},
    {,},{,},{,},{,},
    {,},{,},{,}};

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

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

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

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

    if(R==L) return;

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

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

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

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

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

    return temp;
}

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

    if(L==R)
        return L;

    int M=(L+R)/;

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

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

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

        build_tree(,N,);

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

        las=;

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

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

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

            temp1=update(K,,N,);

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

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

    return ;
}