Eddy's research I
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5050 Accepted Submission(s): 3027
Problem Description Eddy's interest is very extensive, recently he is interested in prime number. Eddy discover the all number owned can be divided into the multiply of prime number, but he can't write program, so Eddy has to ask intelligent you to help him, he asks you to write a program which can do the number to divided into the multiply of prime number factor .
Input The input will contain a number 1 < x<= 65535 per line representing the number of elements of the set.
Output You have to print a line in the output for each entry with the answer to the previous question.
Sample Input 11 9412
Sample Output 11 2*2*13*181
Author eddy
Recommend JGShining 两种做法： 代码：
1 #include<iostream>
2 using namespace std;
3 const int maxn=65540;
4 int arr[maxn]={2,3,5,7};
5 void prime()
6 {
7 bool flag;
8 int k=4;
9 for(int i=11;i<maxn;i++)
10 {
11 flag=true;
12 for(int j=2;j*j<=i;j++)
13 {
14 if(i%j==0)
15 {
16 flag=false;
17 }
18 }
19 if(flag)arr[k++]=i;
20 }
21 }
22 int main()
23 {
24 int n,count;
25 prime();
26 while(cin>>n)
27 {
28 count=0;
29 while(n!=1)
30 {
31 for(int i=0;n!=1;i++)
32 {
33 if(n%arr[i]==0)
34 {
35 if(count++==0)
36 cout<<arr[i];
37 else
38 cout<<"*"<<arr[i];
39 n/=arr[i];
40 break;
41 }
42 }
43 }
44 cout<<endl;
45 }
46 return 0;
47 }
View Code
代码：
1 #include<stdio.h>
2 void show(int a)
3 {
4 int i=1,n=0;
5 while(i<=a&&a!=1)
6 {
7 i++;
8 while(a%i==0)
9 {
10 a/=i; n+=1;
11 printf(n==1?"%d":"*%d",i);
12 }
13
14 }
15 puts("");
16
17 }
18 int main()
19 {
20 int x;
21 while(scanf("%d",&x)!=EOF)
22 show(x);
23 return 0 ;
24 }
View Code
编程是一种快乐，享受代码带给我的乐趣！！！
