PAT--甲 1059 Prime Factors

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甲 1059 Prime Factors (题解)

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

输入格式:

Each input file contains one test case which gives a positive integer N in the range of long int.

输出格式:

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.

1
97532468

输出样例:

1
97532468=2^2*11*17*101*1291

思路:

1.先用素数打表

2.用结构体存储质因子及个数

我的代码如下:

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#include<bits/stdc++.h>
bool prime[100010]={0};
int p[100010]={0};
int find()		//求素数表
{
	int count=0;
	for(int i=2;i<100010;i++)
	{
		if(prime[i]==false)
		{
			p[count++]=i;
			for(int j=2*i;j<100010;j=j+i)
				prime[j]=true;
		}
	}
	return count;
}
struct factor{
	int x;
	int cnt;
}f[10];
int main()
{
	int n,num=0;
	scanf("%d",&n);
	int count=find();
	int sn=(int)sqrt(1.0*n);
	if(n==1)
		printf("1=1\n");
	else
	{
		printf("%d=",n);
		for(int i=0;i<count&&p[i]<=sn;i++)
		{
			if(n%p[i]==0)
			{
				int c=0;
				f[num].x=p[i];
				while(n%p[i]==0)
				{
					c++;
					n=n/p[i];
				}
				f[num].cnt=c;
				num++;
			}
			if(n==1)		//结束循环条件
				break;
		}
		if(n!=1)	//如果有大于根号n的质因子
		{
			f[num].x=n;
			f[num].cnt=1;
			num++;
		}
		for(int i=0;i<num;i++)
		{
			if(i>0)
				printf("*"); 
			printf("%d",f[i].x);
			if(f[i].cnt!=1)
				printf("^%d",f[i].cnt);
		}	
		printf("\n");
	}
	return 0;
}
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