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?

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.

 1 #include <iostream>
 2 //#include<bits/stdc++.h>
 3 #include <stack>
 4 #include <queue>
 5 #include <cstdio>
 6 #include <cstring>
 7 #include <algorithm>
 8 using namespace std;
 9 typedef long long ll;
10 typedef unsigned long long ull;
11 const int MAX=5e5+5;
12 const int antiprime[]={1,2,4,6,12,24,36,48,60,120,180,240,360,720,840,
13                        1260,1680,2520,5040,7560,10080,15120,20160,25200,
14                        27720,45360,50400,55440,83160,110880,166320,221760,
15                        277200,332640,498960,554400,665280
16                       };
17 
18 const int factorNum[]={1,2,3,4,6,8,9,10,12,16,18,20,24,30,32,36,40,48,60,
19                        64,72,80,84,90,96,100,108,120,128,144,160,168,180,
20                        192,200,216,224
21                       };
22 struct node
23 {
24     char name[10];
25     int num;
26 }ren[MAX];
27 int st[10*MAX];
28 void init(int l,int r,int k)
29 {
30     st[k]=r-l;
31     if(l+1==r)
32         return;
33     init(l,(l+r)/2,2*k);
34     init((l+r)/2,r,2*k+1);
35 }
36 void push(int k)
37 {
38     if(k<=0)
39         return;
40     st[k]--;
41     push(k/2);
42 }
43 int update(int dis,int l,int r,int k)
44 {
45     if(l+1==r)
46     {
47         push(k);
48         return l;
49     }
50     if(dis<=st[2*k])
51         return update(dis,l,(l+r)/2,2*k);
52     else
53         return update(dis-st[2*k],(l+r)/2,r,2*k+1);
54 }
55 int main()
56 {
57     int n,k;//n为初始人数，k为每次该出去的人的相对序号（即此时剩的人编上1——x时的序号）
58     int cnt;
59     int i;
60     int pos;//该出去的人的真实序号（初始时这个人的编号）
61     while(~scanf("%d %d",&n,&k))
62     {
63         init(1,n+1,1);
64         for(i=1;i<=n;i++)
65         {
66             scanf("%s %d",ren[i].name,&ren[i].num);
67         }
68         cnt=0;
69         while(n>=antiprime[cnt])
70             cnt++;
71         cnt--;
72         pos=0;
73         ren[0].num=0;
74         for(i=1;i<=antiprime[cnt];i++)
75         {
76             if(ren[pos].num>0)
77                 k=((k+ren[pos].num-2)%st[1]+st[1])%st[1]+1;//此时st[1]个人，k值为1——st[1]，不能为0，故进行此操作
78             else
79                 k=((k+ren[pos].num-1)%st[1]+st[1])%st[1]+1;
80             pos=update(k,1,n+1,1);
81         }
82         printf("%s %d\n",ren[pos].name,factorNum[cnt]);
83     }
84 }