PAT (Advanced Level) 1059. Prime Factors (25) 求所有质因子

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

求出所有质因子存到结果数组，输出控制。

/*2015.7.27cyq*/
#include <iostream>
#include <vector>
using namespace std;

bool isprime(int n){
if(n<=1)
return false;
if(n==2||n==3)
return true;
for(int i=2;i*i<=n;i++)
if(n%i==0)
return false;
return true;
}
int main(){
long N;
cin>>N;
if(N==1){
cout<<"1=1";
return 0;
}
vector<int> result;//结果数组，存储所有质因子
long tmp=N;
while(tmp!=1){
for(int i=2;i<=tmp;i++){
if(isprime(i)&&(tmp%i==0)){
result.push_back(i);
tmp/=i;
break;
}
}
}

cout<<N<<"=";
auto it=result.begin();
cout<<*it;
long pre=*it;
it++;
int count=1;
while(it!=result.end()){
while(it<result.end()&&*it==pre){
count++;
it++;
}
//it指向一个新的数或result.end()
if(count>1){
cout<<"^"<<count;
count=1;
}else{
cout<<"*"<<*it;
pre=*it;
it++;
}
}
return 0;
}

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