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 *…*p_m^k_m.

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 *…*p_m^k_m, where p_i's are prime factors of N in increasing order, and the exponent k_i is the number of p_i -- hence when there is only one p_i, k_i is 1 and must NOT be printed out.

Sample Input:

97532468

Sample Output:

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

总觉得这道题目有点问题，要是输入的大数就是一个素数呢！肯定会超时啊！！！！晕~~~

思路：找出素数列表，依次与N判断即可

AC代码：

#include <iostream>
#include <vector>
#include <math.h>
using namespace std;
struct Node {
    long value;
    int countN;
};
vector<Node> arr(70);
vector<long> primeArr;
long N;

long isPrime(long n){
    if(n==2)
        return 1;
    for(long i=2;i<=sqrt(n);i++){
        if(n%i==0){
            return 0;
        }
    }
    return 1;
}
void findPrime(long n){
    for(long i=2;i<=n;i++){
        if(isPrime(i)){
            primeArr.push_back(i);
        }
    }
}
int main()
{
    cin>>N;
    if(N==1){
        cout<<"1=1";
        return 0;
    }
    long cp = N;
    findPrime(10000);
    int index = 0;
    for(long i=0;i<primeArr.size();i++){
        if(N%primeArr[i]==0){
            if(index==0){
                arr[index].value = primeArr[i];
                arr[index].countN++;
            }else{
                if(primeArr[i]==arr[index].value){
                    arr[index].countN++;
                }else{
                    arr[++index].value = primeArr[i];
                    arr[index].countN++;
                }
            }
            N /= primeArr[i];
            i--;
            if(N==1){
                break;
            }
        }else{
            if(arr[index].countN>0){
                index++;
            }
        }
    }
    cout<<cp<<"=";
    for(int i=0;i<index;i++){
        if(arr[i].countN==1){
            cout<<arr[i].value<<"*";
        }
        if(arr[i].countN>1){
            cout<<arr[i].value<<"^"<<arr[i].countN<<"*";
        }
    }
    if(arr[index].countN==1){
            cout<<arr[index].value;
        }
        if(arr[index].countN>1){
            cout<<arr[index].value<<"^"<<arr[index].countN;
        }
    return 0;
}