Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1k1×p2k2×⋯×pmkm.
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
代码
#include<cstdio>
#include<cmath>
#include<vector>
using namespace std;
typedef long long LL;
vector<int> factor;
vector<int> index;
void Divide(LL N)
{
int Num,Temp=N;
//质因数分解算法
for(int i=2;i*i<=N;i++)
{
if(N%i==0)
{
factor.push_back(i);
Num=0;
while(N%i==0)
{
N/=i;
Num+=1;
}
index.push_back(Num);
}
}
if(N>1)
{
factor.push_back(N);
index.push_back(1);
}
int Len=factor.size();
printf("%d=",Temp);
for(int i=0;i<Len;i++)
{
if(i==0)
{
if(index[i]!=1)
printf("%d^%d",factor[i],index[i]);
else
printf("%d",factor[i]);
}
else
{
if(index[i]!=1)
printf("*%d^%d",factor[i],index[i]);
else
printf("*%d",factor[i]);
}
}
return ;
}
int main()
{
LL N;
scanf("%lld",&N);
if(N==1)
printf("1=1");
else
Divide(N);
return 0;
}