PAT 1059 Prime Factors (25)

最新推荐文章于 2025-08-12 09:57:02 发布

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本文介绍了一种质因数分解算法的实现方法，并通过C++代码详细展示了如何将一个正整数N分解为其所有质因数的形式。该算法能够有效地处理在long int范围内的任意正整数。

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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

解答：本题应该算是水题了，只不过要注意N为1的情形。如果N>1，那么可以遍历寻找因子，题目变得简单直观。我也看了其他人的解法，还是觉得自己的解法是很优秀的，哈哈。

AC代码如下：

#include<iostream>
#include<cstdio>
#include<vector>
#include<cmath>

using namespace std;

typedef struct{
	int val;
	int count;
}prime;

int main()
{
	long int N, C;
	vector<prime> primes;
	
	scanf("%d", &N);
	C = N;
	//如果N是1,那么直接输出结果即可。 
	if( N == 1)
	{
		printf("1=1\n"); return 0;
	}
	//寻找因子 
	for(int i = 2; i <= N; ++i)
	{
		prime tmp; tmp.val = i; tmp.count = 0;
		while(N % i == 0 && N != 0)
		{
			tmp.count++;
			N /= i;
		}
		if(tmp.count != 0) primes.push_back(tmp);
		if(N == 0) break;
	}
	//输出结果 
	printf("%d=", C);
	for(int i = 0; i < primes.size(); ++i)
	{
		if(i == 0) 
		{
			if(primes[i].count == 1)
			{
				printf("%d", primes[i].val);
			}
			else
			{
				printf("%d^%d", primes[i].val, primes[i].count);
			}
		}
		else
		{
			if(primes[i].count == 1)
			{
				printf("*%d", primes[i].val);
			}
			else
			{
				printf("*%d^%d", primes[i].val, primes[i].count);
			}
		}
		
	}
	return 0;
}