1059. Prime Factors (25)

最新推荐文章于 2022-05-16 14:45:04 发布

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最新推荐文章于 2022-05-16 14:45:04 发布

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分类专栏： # 思维技巧 # 暴力 # pat甲级

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本文介绍了一个用于将正整数分解为其质因数的算法，并提供了完整的C++代码实现。该算法能够有效地找出输入整数的所有质因数，并以特定格式输出这些因数及其对应的指数。

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1059. Prime Factors (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

HE, Qinming

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

code:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;
typedef long long ll;
struct node{
    ll factor;
    int e;
}f[1000];
bool judge(ll a){
    ll i;
    ll m = sqrt(a);
    for(i = 2; i <= m; i++){
        if(a % i == 0) return 0;
    }
    return 1;
}
int main(){
    ll n;
    int cnt = 0;
    cin >> n;
    ll tmp = n;
    if(n == 1){
        cout << "1=1" << endl;
        return 0;
    }
    for(int i = 2; i < 100000; i++){
        if(n % i == 0){
            int e = 0;
            while(n % i == 0){
                n /= i;
                e++;
            }
            f[cnt].factor = i;
            f[cnt++].e = e;
        }
    }
    cout << tmp << "=";
    for(int i = 0; i < cnt; i++){
        if(i == 0){
            cout << f[i].factor;
            if(f[i].e > 1) cout << "^" << f[i].e;
        }
        else{
            cout << "*" << f[i].factor;
            if(f[i].e > 1)  cout << "^" << f[i].e;
        }
    }
    if(n == tmp) cout << n << endl;
    else if(n != 1) cout << "*" << n << endl;
    return 0;
}