Bizon the Champion isn‘t just friendly, he also is a rigorous coder.
Let‘s define function f(a), where a is a sequence of integers. Function f(a) returns the following sequence: first all divisors of a1 go in the increasing order, then all divisors of a2 go in the increasing order, and so on till the last element of sequence a. For example, f([2, 9, 1]) = [1, 2, 1, 3, 9, 1].
Let‘s determine the sequence Xi, for integer i (i ≥ 0): X0 = [X] ([X] is a sequence consisting of a single number X), Xi = f(Xi - 1) (i > 0). For example, at X = 6 we get X0 = [6], X1 = [1, 2, 3, 6], X2 = [1, 1, 2, 1, 3, 1, 2, 3, 6].
Given the numbers X and k, find the sequence Xk. As the answer can be rather large, find only the first 105 elements of this sequence.
A single line contains two space-separated integers — X (1 ≤ X ≤ 1012) and k (0 ≤ k ≤ 1018).
Print the elements of the sequence Xk in a single line, separated by a space. If the number of elements exceeds 105, then print only the first 105 elements.
6 1
1 2 3 6
4 2
1 1 2 1 2 4
10 3
1 1 1 2 1 1 5 1 1 2 1 5 1 2 5 10
1 #include<stdio.h> 2 #include<map> 3 #include<algorithm> 4 #define MAXN 20000000 5 using namespace std; 6 typedef long long LL; 7 LL ys[9000];int tot=0;int tt2=0; 8 int tail[9000];int head[9000]; 9 int aft[MAXN];LL p[MAXN]; 10 LL n,k; 11 map<LL,int>bh; 12 void line(int j,int i) 13 { 14 tt2++;if(!tail[j])head[j]=tt2;p[tt2]=i;aft[tail[j]]=tt2;tail[j]=tt2; 15 } 16 void init() 17 { 18 for(LL i=1;i*i<=n;i++) 19 if(n%i==0) 20 { 21 ys[++tot]=i; 22 if(i*i!=n) 23 ys[++tot]=n/i; 24 } 25 sort(ys+1,ys+1+tot); 26 for(int i=1;i<=tot;i++)bh[ys[i]]=i; 27 for(int i=1;i<=tot;i++) 28 for(int j=1;j<=i;j++) 29 if(ys[i]%ys[j]==0)line(i,j); 30 } 31 int dfs(int now,LL dep,int need) 32 { 33 if(ys[now]==1) 34 { 35 printf("1 "); 36 return 1; 37 } 38 if(dep==k) 39 { 40 int us=need; 41 for(int u=head[now];u&&need;need--,u=aft[u]) 42 printf("%I64d ",ys[p[u]]); 43 return us-need; 44 } 45 int us=need; 46 for(int u=head[now];u&&need;need-=dfs(p[u],dep+1,need),u=aft[u]); 47 return us-need; 48 } 49 int main() 50 { 51 scanf("%I64d%I64d",&n,&k);if(k>100000)k=100000; 52 init(); 53 if(!k) 54 { 55 printf("%I64d\n",n); 56 return 0; 57 } 58 dfs(bh[n],1,100000); 59 return 0; 60 }
Codeforces Round #256 (Div. 2) E Divisors,布布扣,bubuko.com
Codeforces Round #256 (Div. 2) E Divisors
原文：http://www.cnblogs.com/qscqesze/p/3852510.html
