FUzhou 1607 Greedy division---因子个数问题。

Problem 1607 Greedy division

http://acm.fzu.edu.cn/problem.php?pid=1607

Accept: 402    Submit: 1463
Time Limit: 1000 mSec    Memory Limit : 32768 KB

Problem Description

Oaiei has inherited a large sum of wealth recently; this treasure has n pieces of golden coins. Unfortunately, oaiei can not own this wealth alone, and he must divide this wealth into m parts (m>1). He can only get one of the m parts and the m parts have equal coins. Oaiei is not happy for only getting one part of the wealth. He would like to know there are how many ways he can divide the wealth into m parts and he wants as many golden coins as possible. This is your question, can you help him?

Input

There are multiply tests, and that will be 500000. For each test, the first line is a positive integer N(2 <= N <= 1000000), N indicates the total number of golden coins in the wealth.

Output

For each test, you should output two integer X and Y, X is the number of ways he can divide the wealth into m parts where m is large than one, Y is the number of golden coins that he can get from the wealth.

Sample Input

5 6 8

Sample Output

1 1 3 3 3 4

Hint

There are huge tests, so you should refine your algorithm.

Source

FOJ月赛-2008年5月

根据数论公式：

 

 1 /*
 2 题意：一个人，想知道财产的分法有多少种，(财产必须平方的前提)
 3       他想分得越多越好。
 4       求出N M。
 5 用欧拉的模板稍微改变一下。
 6 N，容易解决。
 7 M的求解不是
 8 if(M&1) M=1;
 9 else M=N/2;
10 显然15的时候是 5
11 实际上就是 num=max{  N/(枚举每个素因子) }
12 再加一个判断 if(num==N) num=1;
13 
14 */
15 
16 #include<stdio.h>
17 #include<stdlib.h>
18 
19 int Num_Euler(int n,int *m)
20 {
21     int num=1,k,i;
22     for(i=2;i*i<=n;i++)
23     if(n%i==0)
24     {
25         k=1;
26         while(n%i==0)
27         {
28             if(n/i>*m)
29             *m=n/i;
30             k++;
31             n=n/i;
32         }
33         num=num*k;
34     }
35     if(n!=1)
36     {
37         num=num*2;
38         if(n>*m)
39         *m=n;
40     }
41     return num;
42 }
43 
44 int main()
45 {
46     int n,k,m;
47     while(scanf("%d",&n)>0)
48     {
49         m=-1;
50         k=Num_Euler(n,&m);
51         if(m==n) m=1;//加的判断,当n是一个素数的时候，在Num_Euler(n,&m)最后就会得到m==n 但是题目要求至少分成2份啊，所以为1
52         printf("%d %d\n",k-1,m);//K-1个，不包含自己.
53     }
54     return 0;
55 }