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

 1 #include <iostream>
 2 #include <cmath>
 3 const int maxn = 100010;
 4 bool isprime(int n)
 5 {//判断n是否为素数
 6     if (n == 1)
 7         return false;
 8     int sqr = (int)sqrt(1.0*n);
 9     for (int i = 2; i <= sqr; i++)
10     {
11         if (n % i == 0)
12             return false;
13     }
14     return true;
15 }
16 int prime[maxn], pNum = 0;
17 void FindPrime()
18 {//求素数表
19     for (int i = 1; i < maxn; i++)
20     {
21         if (isprime(i) == true)
22             prime[pNum++] = i;
23     }
24 }
25 struct factor 
26 {
27     int x,cnt;//x为质因子，cnt为其个数
28 }fac[10];
29 
30 int main()
31 {
32     FindPrime();//此句必须记得写
33     int n, num = 0;//num为n的不同质因子的个数
34     scanf("%d", &n);
35     if (n == 1)
36         printf("1=1");//特判1的情况
37     else
38     {
39         printf("%d=", n);
40         int sqr = (int)sqrt(1.0*n);//n的根号
41         //枚举根号n以内的质因子
42         for (int i = 0; i < pNum&&prime[i] <= sqr; i++)
43         {
44             if (n%prime[i] == 0)//如果prime[i]是n的因子
45             {
46                 fac[num].x = prime[i];//记录该因子
47                 fac[num].cnt = 0;
48                 while (n%prime[i] == 0)
49                 {//计算出质因子prime[i]的个数
50                     fac[num].cnt++;
51                     n /= prime[i];
52                 }
53                 num++;//不同质因子个数加1
54             }
55             if (n == 1)
56                 break;//及时退出循环，节省点时间
57         }
58         if (n != 1)
59         {//如果无法被根号n以内的质因子除尽
60             fac[num].x = n;//那么一定有一个大于根号n的质因子
61             fac[num++].cnt = 1;
62         }
63         //按格式输出结果
64         for (int i = 0; i < num; i++)
65         {
66             if (i > 0)
67                 printf("*");
68             printf("%d", fac[i].x);
69             if (fac[i].cnt > 1)
70                 printf("^%d", fac[i].cnt);
71         }
72     }
73     return 0;
74 }