poj 2886 线段树+反素数

Who Gets the Most Candies?

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 12744		Accepted: 3968
Case Time Limit: 2000MS

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?

Input

There are several test cases in the input. Each test case starts with two integers N (0 < N ≤ 500,000) and K (1 ≤ K ≤ N) on the first line. The next N lines contains the names of the children (consisting of at most 10 letters) and the integers (non-zero with magnitudes within 10⁸) on their cards in increasing order of the children’s numbers, a name and an integer separated by a single space in a line with no leading or trailing spaces.

Output

Output one line for each test case containing the name of the luckiest child and the number of candies he/she gets. If ties occur, always choose the child who jumps out of the circle first.

Sample Input

4 2
Tom 2
Jack 4
Mary -1
Sam 1

Sample Output

Sam 3

/*
poj 2886 线段树+反素数
反素数：
（1）给定一个数，求一个最小的正整数，使得的约数个数为
（2）求出中约数个数最多的这个数

反素数以前也大概学习过，但是做这个题的时候并没有想到优化方法
问题很严重- -

给你n个数围成一圈，然后每个人对应一个数字
如果x是正数，则下一个人是左边第x个。如果x为负数，则下一个人是
右边第-x个

大致思路是没啥问题，但是开始是将所有数进行预处理，求出第i个数
的f[i]，然后同过线段树判断左右两边的剩余人数，然后TLE

后来发现别人都是先反素数打表，rprim[0]表示有rprim[1]个因子时的最
大值,然后便能1-n中f[]最大值p的位置。然后只需从1模拟到p求出这时对
应的人即可

hhh-2016-03-23 23:50:13
*/
#include <algorithm>
#include <cmath>
#include <queue>
#include <iostream>
#include <cstring>
#include <map>
#include <cstdio>
#include <vector>
#include <functional>
#define lson (i<<1)
#define rson ((i<<1)|1)
using namespace std;
typedef long long ll;
const int maxn = 500550;
int rprim[35][2] =
{
    498960,200,332640,192,277200,180,221760,168,166320,160,
    110880,144,83160,128,55440,120,50400,108,45360,100,
    27720,96,25200,90,20160,84,15120,80,10080,72,
    7560,64,5040,60,2520,48,1680,40,1260,36,
    840,32,720,30,360,24,240,20,180,18,
    120,16,60,12,48,10,36,9,24,8,
    12,6,6,4,4,3,2,2,1,1
};
struct node
{
    int l,r;
    int num;
    int mid()
    {
        return ((l+r)>>1);
    };
} tree[maxn<<2];

void update_up(int i)
{
    tree[i].num = tree[lson].num + tree[rson].num;
}

void build(int i,int l,int r)
{
    tree[i].l = l,tree[i].r = r;

    if(l == r)
    {
        tree[i].num = 1;
        return ;
    }
    int mid = tree[i].mid();
    build(lson,l,mid);
    build(rson,mid+1,r);
    update_up(i);
}

void update_down(int i)
{

}
int cur;
void Insert(int i,int k)
{
    if(tree[i].l ==  tree[i].r)
    {
        cur = tree[i].l;
        tree[i].num = 0;
        return ;
    }
    update_down(i);
    int mid = tree[i].mid();
    if(k <= tree[lson].num)
        Insert(lson,k);
    else
        Insert(rson,k-tree[lson].num);
    update_up(i);
}

int query(int i,int l,int r)
{
    if(tree[i].l >= l && tree[i].r <= r)
        return tree[i].num;
    update_down(i);
    int mid = tree[i].mid();
    int all = 0;
    if(l <= mid)
        all += query(lson,l,r);
    if(r > mid)
        all += query(rson,l,r);
    return all;
}

char nam[maxn][13];
int a[maxn];

int main()
{
    int k,n,t,m,nex;
    while(scanf("%d%d",&n,&k) != EOF)
    {
        t = 0;
        while(n < rprim[t][0])
            t++;
        int x = rprim[t][0];
        for(int i = 1; i <= n; i++)
        {
            scanf("%s%d",nam[i],&a[i]);
        }
        build(1,1,n);
        cur = k,m = n;
        for(int i = 1; i < x; i++)
        {
            //cout << cur << endl;
            Insert(1,cur);
           //cout << cur <<endl;
            nex = a[cur];
            m--;
            if(nex%m == 0)
            {
                if(nex < 0) nex = 1;
                else nex = m;
            }
            else
            {
                nex %= m;
                if(nex < 0) nex += (m+1);
            }
            int leftnum = query(1,1,cur);
            int rightnum = m - leftnum;
            if(nex <= rightnum) cur = leftnum + nex;
            else cur = nex-rightnum;
        }
        Insert(1,cur);
        printf("%s %d\n",nam[cur],rprim[t][1]);
    }
    return 0;
}