PAT甲级 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 *…*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
思路:

1.先构建质数表,根据N的大小,把1~sqrt(N)之间的质数找出来;

2.根据质数表,从小到大把N的因子找出,这里利用map比较好做;

3.注意N=1的情况

代码:

#include
#include
#include
#include
using namespace std;

int main()
{
	//ifstream cin;
	//cin.open("case1.txt");

	long N;
	cin >> N;

	if (N == 1)
		printf("1=1\n");
	else
	{
		//构建质数表
		bool *prime = (bool *)malloc(sizeof(bool) * sqrt(N));
		memset(prime, 0, sizeof(bool)*sqrt(N));

		int j, i = 2;
		for (; i <= sqrt(N); ++i)
		{
			if (!prime[i])
			{
				for (j = 2 * i; j <= sqrt(N); j += i)
				{
					prime[j] = 1;
				}
			}
		}
		//根据质数表进行分解N
		long NN = N;
		map p;
		for (i = 2; i <= N;)
		{
			if ((!prime[i]) && (N%i == 0))
			{
				//遇到质数,且该质数是N的因数
				p[i]++;
				N /= i;
			}
			else
			{
				//跳过
				++i;
			}
		}
		//output
		map::iterator iter = p.begin();
		printf("%ld=", NN);
		while (1)
		{
			printf("%d", iter->first);
			if (iter->second > 1)
			{
				printf("^%d", iter->second);
			}

			iter++;
			if (iter != p.end())
			{
				printf("*");
			}
			else
			{
				break;
			}
		}

		printf("\n");
	}
    system("pause");
	return 0;
}


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