PAT甲级——A1059 Prime Factors

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 <iostream>
 #include <cmath>
 const int maxn = ;
 bool isprime(int n)
 {//判断n是否为素数
     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 FindPrime()
 {//求素数表
     for (int i = ; i < maxn; i++)
     {
         if (isprime(i) == true)
             prime[pNum++] = i;
     }
 }
 struct factor
 {
     int x,cnt;//x为质因子，cnt为其个数
 }fac[];

 int main()
 {
     FindPrime();//此句必须记得写
     int n, num = ;//num为n的不同质因子的个数
     scanf("%d", &n);
     if (n == )
         printf("1=1");//特判1的情况
     else
     {
         printf("%d=", n);
         int sqr = (int)sqrt(1.0*n);//n的根号
         //枚举根号n以内的质因子
         for (int i = ; i < pNum&&prime[i] <= sqr; i++)
         {
             if (n%prime[i] == )//如果prime[i]是n的因子
             {
                 fac[num].x = prime[i];//记录该因子
                 fac[num].cnt = ;
                 while (n%prime[i] == )
                 {//计算出质因子prime[i]的个数
                     fac[num].cnt++;
                     n /= prime[i];
                 }
                 num++;//不同质因子个数加1
             }
             if (n == )
                 break;//及时退出循环，节省点时间
         }
         if (n != )
         {//如果无法被根号n以内的质因子除尽
             fac[num].x = n;//那么一定有一个大于根号n的质因子
             fac[num++].cnt = ;
         }
         //按格式输出结果
         for (int i = ; i < num; i++)
         {
             if (i > )
                 printf("*");
             printf("%d", fac[i].x);
             if (fac[i].cnt > )
                 printf("^%d", fac[i].cnt);
         }
     }
     return ;
 }