pat 1059

1059. Prime Factors (25)

时间限制

50 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

HE, Qinming

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

#include <stdio.h>

int priNum = 0;
int ansPrime[10010];
int mark[100010];
int outP[1010], outZ[1010];

void Prime()
{
	int i, j;

	for(i = 2; i <= 100000; i++)
	{
		if(mark[i] == 0)
			ansPrime[priNum++] = i;

		for(j = i + i; j <= 100000; j += i)
			mark[j] = 1;
	}
}

int main()
{
	int n, i, count, tmp;

	Prime();

	while(scanf("%d", &n) != EOF)
	{
		i = 0;
		count = 0;

		tmp = n;

		if(n == 1)
		{
			printf("1=1\n");
			continue;
		}

		while(n != 1)
		{
			if(n % ansPrime[i] == 0)
			{
				count++;

				while(n % ansPrime[i] == 0)
				{
					outP[count] = ansPrime[i];
					outZ[count]++;
					n = n / ansPrime[i];
				}
			}
			i++;
		}

		printf("%d=", tmp);
		for(i = 1; i <= count; i++)
		{
			
			printf("%d", outP[i]);

			if(outZ[i] > 1)
			{
				printf("^%d", outZ[i]);
			}

			if(i != count)
				printf("*");
			else
				printf("\n");
		}
	}
	return 0;
}