PAT1059:Prime Factors

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

思路

  按指定格式打印一个数的所有素（质）数因子。
  1.先初始化建立一个素数表。
  2.从2开始，不断用素数i除输入的数num，能够整除说明满足条件，用cnt记录被整除的次数。
  3.对于第一个除数的输出进行判断来决定在除数前面是否加"*"。
  4.cnt > 2就在当前除数后加"^cnt"。
  5.重复2.3.4直至num < 2。
  注：对于1要进行特殊处理，不然输出为"1=",正确输出应该是"1=1"。
代码

#include<iostream>
#include<vector>
using namespace std;

vector<int> isPrime(666666,1);

void Init()
{
    for(int i = 2; i * i < 666666; i++)
    {
        for(int j = 2; j*i < 666666; j++)
        {
            isPrime[i * j] = 0;
        }
    }
}

int main()
{
    Init();
    long long num;
    cin >> num;
    cout << num <<"=";
    if(num == 1)
        cout << 1;
    bool first = true;
    for(int i = 2; num >= 2; i++)
    {
        int cnt = 0;
        if(isPrime[i] == 1 && num % i == 0)
        {
            while(num % i == 0)
            {
                cnt++;
                num = num/i;
            }
            if(first)
                first = false;
            else
                cout << "*";
            cout << i;
            if(cnt > 1)
                cout << "^" << cnt;
        }
    }
}