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 = 100010;
bool isprime(int n)
{//判断n是否为素数
    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 FindPrime()
{//求素数表
    for (int i = 1; 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[10];

int main()
{
    FindPrime();//此句必须记得写
    int n, num = 0;//num为n的不同质因子的个数
    scanf("%d", &n);
    if (n == 1)
        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 = 0; i < pNum&&prime[i] <= sqr; i++)
        {
            if (n%prime[i] == 0)//如果prime[i]是n的因子
            {
                fac[num].x = prime[i];//记录该因子
                fac[num].cnt = 0;
                while (n%prime[i] == 0)
                {//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 == 1)
                break;//及时退出循环，节省点时间
        }
        if (n != 1)
        {//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 = 1;
        }
        //按格式输出结果
        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;
}