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1059. Prime Factors (25)


1059. Prime Factors (25)

时间限制



50 ms



内存限制



65536 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 *…*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


这题和之前的一题很像。http://xujiayu317.blog.163.com/blog/static/2547520920149218497214/


但是这题过程中一直有个结果错误。看了那些ac的代码。发现是要当num=1时要1=1  改了一下就过了(但是1不是质数吧,这属于特殊条件)。


评测结果

时间

结果

得分

题目

语言

用时(ms)

内存(kB)

用户

7月17日 20:38

答案正确

​​25​​

​​1059​​

​​C++ (g++ 4.7.2)​​

1

308

​​datrilla​​

测试点

测试点

结果

用时(ms)

内存(kB)

得分/满分

0

答案正确

1

180

15/15

1

答案正确

1

308

4/4

2

答案正确

1

180

2/2

3

答案正确

1

180

2/2

4

答案正确

1

180

2/2

#include<iostream>
#include<math.h>
using namespace std;
class PrimeFactors
{
private:
long int e,m;
PrimeFactors* next;
public:
void printpf();
void line(long int ee,long int mm);
PrimeFactors*head,*sail;
PrimeFactors()
{
head=NULL;
sail=NULL;
}
};
void PrimeFactors::printpf()
{
cout<<head->e<<"=";
sail=head;
head=head->next;
delete sail;
while(head->next)
{
cout<<head->e;
if(head->m!=1)
{
cout<<"^"<<head->m;
}
cout<<"*";sail=head;
head=head->next;
delete sail;
}
cout<<head->e;
if(head->m!=1)
{
cout<<"^"<<head->m;
}
delete head;
cout<<endl;
head=NULL;
sail=NULL;
}
void PrimeFactors::line(long int ee,long int mm)
{
PrimeFactors*p;
if(head)
{
p=new PrimeFactors;
p->e=ee;
p->m=mm;
sail->next=p;
sail=p;
}
else
{
head=new PrimeFactors;
head->e=ee;
head->m=0;
sail=head;
}
sail->next=NULL;
}
int main()
{
PrimeFactors pf;
long int num,add,count;
cin>>num;
if(num>1)
{
pf.line(num,0);
add=2;
count=0;
while(num%add==0)
{
num/=add;
count++;
}
if(count!=0)
{
pf.line(add,count);
}
add=3;
while(num>2&&add<=sqrt(num))
{
count=0;
while(num%add==0)
{
count++;
num/=add;
}
if(count!=0)
{
pf.line(add,count);
}
add+=2;
}
if(num>2)
pf.line(num,1);
pf.printpf();
}
else cout<<num<<"="<<num<<endl;
return 0;
}



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