1059 Prime Factors（25 分）

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<cstdio>
#include<cmath>
const int maxn = ;

bool is_prime(int n){
    if(n == ) return false;
    int sqr = (int)sqrt(1.0*n);
    for(int i = ; i <= sqr; i++){
        if(n % i == ) return false;
    }
    return true;
}

int prime[maxn],pNum = ;
void Find_prime(){
    for(int i =  ; i < maxn; i++){
        if(is_prime(i) == true){
            prime[pNum++] = i;
        }
    }
}

struct facot{
    int x,cnt;
}fac[];
int main(){
    Find_prime();
    int n;
    scanf("%d",&n);
    int num = ;
    if(n == ) printf("1=1");
    else{
        printf("%d=",n);
        int sqr = (int)sqrt(1.0*n);
        //printf("prime[0]");
        for(int i = ; i < pNum ; i++){
            //printf("%d",i);
            if(n % prime[i] == ){
                fac[num].x = prime[i];
                fac[num].cnt = ;

            while(n % prime[i] == ){
                fac[num].cnt++;
                n /= prime[i];
            }
            num++;
        }
        if(n == ) break;
    }
    if(n != ){
        fac[num].x = n;
        fac[num].cnt = ;
    }
    //printf("1\n");
    for(int i = ; i < num; i++){
        if(i > ) printf("*");
        printf("%d",fac[i].x);
        if(fac[i].cnt > ) printf("^%d",fac[i].cnt);
    }
}
    return ;
}