A1059. Prime Factors (25)

Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1^k1 * p2^k2 *…*pm^km.

Input Specification:

Each input file contains one test case which gives a positive integer N in the range of long int.

Output Specification:

Factor N in the format N = p1^k1 * p2^k2 *…*pm^km, where pi's are prime factors of N in increasing order, and the exponent ki is the number of pi -- hence when there is only one pi, ki is 1 and must NOT be printed out.

Sample Input:

97532468

Sample Output:

97532468=2^2*11*17*101*1291

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #include <iostream>
 4 #include <string.h>
 5 #include <string>
 6 #include <math.h>
 7 #include <algorithm>
 8 using namespace std;
 9 
10 
11 const int maxn=100000;
12 int prime[maxn],pnum=0;
13 bool p[maxn]={0};
14 void Find_Prime(){
15     for(int i=2;i<maxn;i++)
16     {
17         if(p[i]==false)
18         {
19             prime[pnum++]=i;
20             for(int j=i+i;j<maxn;j+=i)
21             {
22                 p[j]=true;
23             }
24         }
25     }
26 } 
27 
28 struct factor{
29     int x,cnt;
30 }fac[10];
31 int main(){
32    Find_Prime();
33    int n,num=0;
34    scanf("%d",&n);
35    if(n==1){ 
36    printf("1=1");
37    return 0;
38    }else{
39        printf("%d=",n);
40    }
41    
42    
43    int sqr=(int)sqrt(1.0*n);
44    int count=0;
45    for(int i=2;i<=sqr;)
46    {
47        if(n%i==0)
48        {
49            fac[num].x=i;
50            fac[num].cnt=0;
51            while(n%i==0)
52            {
53                fac[num].cnt++;
54                n/=i;
55            }
56            num++;    
57        }
58        
59        count++;
60        i=prime[count];
61    }
62    if(n>sqr)
63    {
64        fac[num].x=n;
65        fac[num++].cnt=1;
66    } 
67    for(int i=0;i<num;i++)
68    {
69        if(i>0)printf("*");
70        printf("%d",fac[i].x);
71        if(fac[i].cnt>1)
72        {
73            printf("^%d",fac[i].cnt);
74        }
75    } 
76     return 0;
77 }

 

posted @ 2015-02-27 16:10  Joilee  阅读(223)  评论(0)  编辑  收藏  举报