题目链接 

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 *…*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 = p₁^k₁ * p₂^k₂ *…*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.（坑爹，这最后一句不是说满足情况不输出，可答案是要输出的，害我吓考虑2个case没过）

Sample Input:

97532468

Sample Output:

97532468=2^2*11*17*101*1291 ---------------------------------------------------华丽的分割线---------------------------------------------------------------------------------------------

1 #include 
     
       2 #include 
      
        3 using namespace std; 4 const int maxn=100005; 5 int prime[maxn],pnum=0; 6 bool p[maxn]={
   0}; 7 void Find_Prime(){ 8     p[0]=p[1]=true; 9     for(int i=2;i
       
        1)47             printf("%d^%d",fac[i].x,fac[i].cnt);48         else49             printf("%d",fac[i].x);50     }51     printf("\n");52 }53 int main()54 {55     Find_Prime();56     int n;57     while(scanf("%d",&n)!=EOF){58         if(n==1)59             printf("1=1\n");60         else{61             PrimeFactor(n);62             Print_fac(n);63         }64 65         /*66         int a=(1<<31)-1;67         cout<
        <