Who Gets the Most Candies?（线段树 + 反素数 ）

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<string.h>
 4 #define lson o << 1 , l , mid
 5 #define rson o << 1 | 1 , mid + 1 , r
 6 using namespace std;
 7 //list anti prime
 8 int s[40] = {1,2,4,6,12,24,36,48,60,120,180,240,360,720,840,1260,1680,2520,5040,7560,10080,15120,20160,25200,27720,45360,50400,55440,83160,110880,166320,221760,277200,332640,498960,500001};
 9 int b[40] = {1,2,3,4,6,8,9,10,12,16,18,20,24,30,32,36,40,48,60,64,72,80,84,90,96,100,108,120,128,144,160,168,180,192,200,1314521};
10 int k , n ;
11 char name[500111][11] ;
12 int val[500111] ;
13 int a[2000100];
14 
15 void init (int o , int l , int r )
16 {
17     a[o] = r - l + 1 ;
18     if ( l == r ) {
19         return ;
20     }
21     int mid = ( l + r ) >> 1 ;
22     init ( lson ) ;
23     init ( rson ) ;
24 }
25 
26 int insert (int x , int o , int l , int r )
27 {
28     a[o]-- ;
29     if ( l == r ) {
30         return l ;
31     }
32     int mid = ( l + r ) >> 1 ;
33     if ( x <= a[o << 1] ) {
34         insert (x , o << 1 , l , mid ) ;
35     }
36     else {
37         insert ( x - a[o << 1] , o << 1 | 1 , mid + 1 , r ) ;
38     }
39 }
40 
41 int main ()
42 {
43   //  freopen ( "a.txt" , "r" , stdin ) ;
44     int i , j , maxn , p , pos ;
45     while (~ scanf ("%d%d" , &n , &k ) ) {
46         i = 0 ;
47         while ( s[i] <= n ) {
48             i++ ;
49         }
50         maxn = b[i - 1] ;
51         p = s[i - 1] ;//p-th leave the team must be the most , and was getting as much Max candy the first team , then
52         //we obtain its number of factor
53         for ( i = 1 ; i <= n ; i++ ) {
54             scanf ( "%s%d" , name[i] , &val[i] ) ;
55         }
56         init (1 , 1 , n ) ;
57         i = n ;
58         while ( p-- ) {//so we can only query p childs
59             n-- ;
60             pos = insert ( k , 1 , 1 , i) ;
61             if ( !n ) {
62                 break ;
63             }
64             if ( val[pos] >= 0 ) {//clockwise
65                 k = ( k - 1 + val[pos] - 1 ) % n + 1 ;//relates to the modular arithmetic , make the code from the begining of ‘0‘ ,
66                 //so start to minus one , as with list Vacancy Segment tree , the obtained ‘k‘ is the new tree i-th location
67             }
68             else {//Anti-clockwise
69                 k = ( ( k - 1 + val[pos] ) % n + n ) % n + 1 ;
70             }
71         }
72         printf ( "%s %d\n" , name[pos] , maxn ) ;
73     }
74     return 0 ;
75 }