1059. Prime Factors (25) PAT 甲级

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本文介绍了一种质因数分解算法的实现方法，通过给定一个正整数N，找到其所有质因数并以特定格式输出。该算法首先构建一个质数表，然后利用这些质数对输入的整数进行分解。

Problem Description

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

传送门

#include<iostream>
#include<math.h>

using namespace std;

#define MAX_N 100010


struct factor{
    int x;
    int cnt;
}fact[10];

bool is_prime(int n){
    if(n==1)    return false;
    int sqr=(int)sqrt(n);

    for(int i=2;i<=sqr;i++){
        if(n%i==0)  return false;
    }
    return true;

}

int prime[MAX_N];
int pNum=0;

void prime_Table(){

    for(int i=1;i<MAX_N;i++){
        if(is_prime(i)){
            prime[pNum++]=i;
        }
    }
}



int main(){
    prime_Table();

    int n;
    int num=0;
    cin>>n;
    if(n==1)
        cout<<"1=1";
    else{
        cout<<n<<"=";
        int sqr=(int)sqrt(n);
        for(int i=0;i<pNum&&prime[i]<=sqr;i++){
            if(n%prime[i]==0){
                fact[num].x=prime[i];
                fact[num].cnt=0;
                while(n%prime[i]==0){
                    fact[num].cnt++;
                    n/=prime[i];
                }
                num++;
            }
            if(n==1)    break;
        }

        if(n!=1)    {
            fact[num].x=n;
            fact[num++].cnt=1;
        }

        for(int i=0;i<num;i++){
            if(i>0) cout<<"*";
            cout<<fact[i].x;
            if(fact[i].cnt>1){
                cout<<"^"<<fact[i].cnt;
            }
        }
    }




}