1059 Prime Factors (25 分)(素数表的建立)

Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p

1

k

1

×p

2

k

2

××p

m

k

m

.

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 = p

1

^k

1

*p

2

^k

2

*…*p

m

^k

m

, where p

i

's are prime factors of N in increasing order, and the exponent k

i

is the number of p

i

-- hence when there is only one p

i

, k

i

is 1 and must NOT be printed out.

Sample Input:

97532468

Sample Output:

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

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
vector prime(500000,1);
int main(){
    for(int i=2;i*i<500000;i++)  //建立素数表
        for(int j=2;j*i<500000;j++)
            prime[i*j]=0;
    long int a;
    cin>>a;
    cout<=2;i++){
        int cnt=0,flag=0;
        while(prime[i]&&a%i==0){
            cnt++;
            a/=i;
            flag=1;
        }
        if(flag){
            if(state) cout<<"*";
            cout<=2)
            printf("^%d",cnt);
    }
    return 0;
}

 

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