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 …pm^km.

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 …pm^km, where pi's are prime factors of N in increasing order, and the exponent ki is the number of pi -- hence when there is only one pi, ki is 1 and must NOT be printed out.

Sample Input:

97532468

Sample Output:

97532468=2^2*11*17*101*1291

题目大意：

给出一个整数，按照从小到大的顺序输出其分解为质因数的乘法算式

分析：

根号int的最大值不会超过50000，先建立个50000以内的素数表，然后从2开始一直判断是否为它的素数，如果是就将a=a/i继续判断i是否为a的素数，判断完成后输出这个素数因子和个数，用state判断是否输入过因子，输入过就要再前面输出。
原文链接：https://blog.csdn.net/liuchuo/article/details/52261852

题解

#include 

using namespace std;
const int maxn=100010;
//判断n是否为素数
bool is_prime(int n)
{
    if(n==1)
        return false;
    for(int i=2; i*i<=n; i++)
    {
        if(n%i==0)
            return false;
    }
    return true;
}
int prime[maxn],pNum=0;
//求素数表
void Find_Prime()
{
    for(int i=1; i1)
            printf("^%d",fac[i].cnt);
    }
    return 0;
}