HDUOJ----Eddy's research I
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 }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 }腾讯云开发者
