PAT甲级1059. Prime Factors (25)

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

#include <vector>#include <iostream>#include <cmath>using namespace std;bool isPrime(long int N){    if(N==1) return false;    if(N==2) return true;    int r=floor(sqrt(N)+0.5);    for(long int i=2;i<=r;i++) {        if(N%i==0) return false;    }    return true;}struct PF{    long int prime;    long int expo;    PF(long int x,long int y):prime(x),expo(y){}};vector<PF> getPF(long int N){    vector<PF> result;    int count=0;    long int num=N;    for(int i=2;i<=N;i++){        if(num<=1) break;//没有这句会超时         int j=0;        int flag=1;        if(isPrime(i)) {            count=0;            while(num%i==0){                num/=i;                count++;            }            if(count>0) {                result.push_back(PF(i,count));            }        }    }    return result;}int main(){    long int N;    cin>>N;    if(N==1) {        cout<<"1=1"<<endl;        return 0;    }    vector<PF> result=getPF(N);    int sz=result.size();    cout<<N<<"=";    if(result[0].expo>1) cout<<result[0].prime<<"^"<<result[0].expo;    else cout<<result[0].prime;    for(long int k=1;k<sz;k++){        if(result[k].expo>1) cout<<"*"<<result[k].prime<<"^"<<result[k].expo;        else cout<<"*"<<result[k].prime;    }    return 0;}