PAT A 1059. Prime Factors (25)

题目

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.

Sample Input:

97532468

Sample Output:

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

 

即求质因数分解，

由小向大扫描，依次取余判断，如果有相应的质因数项，考虑系数。

注意，输入可能是1，输入可能是质数。

 

代码：

#include <iostream>
#include <cmath>
using namespace std;

int main()
{
	long p[65];	//存质因子
	int k[65]={0};	//存系数

	long n;	//输入的数
	cin>>n;
	cout<<n<<"=";
	if(n==1)	//！！！输入可能为1
	{
		cout<<1;
		return 0;
	}

	long i=0;	//记录非重复质因子数量
	long j=2;
	int flag=0;	//获得质因子标志位，用于重新计算sqr
	long sqr=sqrt((double)n)+1;	//求根，计算质因子时需要的上界

	while(n>1)
	{
		if(flag==1)	//重新计算上界
		{
			sqr=sqrt((double)n)+1;
			flag=0;
		}
		for(;j<=sqr;j++)	//从上次的位置开始扫描
		{
			if(n%j==0)	//扫描到一个
			{
				p[i]=j;	//记录
				k[i]=1;
				n/=j;
				while(n%j==0)	//获取系数
				{
					n/=j;
					k[i]++;
				}
				i++;	//刷新数据，为扫描下一个做准备
				j++;
				flag=1;
				break;
			}
		}
		if(j>sqr)	//超界，意味着本身是质数
		{
			p[i]=n;
			k[i]=1;
			i++;
			break;
		}
	}

	if(k[0]==1)	//输出
		cout<<p[0];
	else
		cout<<p[0]<<"^"<<k[0];
	for(j=1;j<i;j++)
	{
		if(k[j]==1)
			cout<<"*"<<p[j];
		else
			cout<<"*"<<p[j]<<"^"<<k[j];
	}

	return 0;
}