描述

传送门：poj-2886

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 $(0 < N \leq 500,000)$ and K $(1 \leq K \leq 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 $10^8$) 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.

Examples

intput
1
2
3
4
5
4 2
Tom 2
Jack 4
Mary -1
Sam 1
output
1
Sam 3

题目大意

有N个人，每个人有一个数$a_i$。从第K个人开始出列，如果他的数$a_i$>0，则他顺时针方向第$(a_i)$个人出列，循环，直到最后一个人出列。

$x$表示在第$x$轮出列,$f(x)$表示$x$的因数个数。求最大的$f(x)$。

思路

线段树维护区间值和，表示这个区间还剩多少人，每次更新就在对应区间-1。

设当前这轮$k$出列，用$k=(k+a_i$)%当前剩余人数，求得下一个$k$的值。

代码

#include<iostream>
#include<algorithm>
#include<string.h>
#include<string>
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
using namespace std;
#define CRL(a) mamset(a,0,sizeof(a))
typedef long long ll;
const int N=5e5+5;
typedef pair<char[12],int > pp;
int tr[N<<2],book[N];
pp people[N];

int Init()		//打表book[i]表示i的因数个数 
{
    for(int i=1;i<=500000;i++)
        for(int j=i;j<=500000;j+=i)
            book[j]++;
}

void Built(int root,int l,int r)
{	
    if(l==r) { tr[root]=1; return;}
    else{
        int mid=l+r>>1;
        Built(root<<1,l,mid);
        Built(root<<1|1,mid+1,r);
        tr[root]=tr[root<<1]+tr[root<<1|1];
    }
}

int Update(int root,int l,int r,int x) 
{
    if(l==r) {
        tr[root]--;
        return l;
    }
    else{
        tr[root]--;
        int mid=l+r>>1;
        if(x<=tr[root<<1])	return Update(root<<1,l,mid,x);		//判断出列的在左边还是右边 
        else   return Update(root<<1|1,mid+1,r,x-tr[root<<1]);
    }
}

int main()
{
    ios::sync_with_stdio(false);
    Init();
    int n,k,ans=0,tem;
    while(cin>>n>>k)
    {
        ans=0;tem=k;
        Built(1,1,n);
        for(int i=1;i<=n;i++)
            cin>>people[i].first>>people[i].second;
        
        for(int i=1;i<=n;i++)
            if(ans<book[i]) ans=book[i];	//找到最大的f(x) 
        
        for(int i=1;i<=n;i++)		//依次出列 
        {
            tem=Update(1,1,n,k);	//第k个人出列 
            if(book[i]==ans)
            {
                cout<<people[tem].first<<" "<<ans<<endl;
                break;
            }
            
            if(people[tem].second>0)  //判断下一个出列的人的位置 
                k=(k-1+people[tem].second-1)%(n-i)+1; 
            else    
                k=((k-1+people[tem].second)%(n-i)+(n-i))%(n-i)+1; 
        }
    }
    return 0;
}