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 = 100010;bool is_prime(int n){if(n == 1) return false;int sqr = (int)sqrt(1.0*n);for(int i = 2; i <= sqr; i++){if(n % i == 0) return false;}return true;
}int prime[maxn],pNum = 0;
void Find_prime(){for(int i = 1 ; i < maxn; i++){if(is_prime(i) == true){prime[pNum++] = i;}}
}struct facot{int x,cnt;
}fac[10];
int main(){Find_prime();int n;scanf("%d",&n);int num = 0;if(n == 1) printf("1=1");else{printf("%d=",n);int sqr = (int)sqrt(1.0*n);//printf("prime[0]");for(int i = 0; i < pNum ; i++){//printf("%d",i);if(n % prime[i] == 0){fac[num].x = prime[i];fac[num].cnt = 0;while(n % prime[i] == 0){fac[num].cnt++;n /= prime[i];}num++;}if(n == 1) break;}if(n != 1){fac[num].x = n;fac[num].cnt = 1;}//printf("1\n");for(int i = 0; i < num; i++){if(i > 0) printf("*");printf("%d",fac[i].x);if(fac[i].cnt > 1) printf("^%d",fac[i].cnt);}
}return 0;
}