1059 Prime Factors

Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p​1​​​k​1​​​​×p​2​​​k​2​​​​×⋯×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 = p​1​​^k​1​​*p​2​​^k​2​​*…*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

题意：

　　给出一个数，要求写出其质因子乘积的形式，相同的质因子用指数表示。

思路：

　　首先打出一个素数表，然后遍历素数表中的素数，如果该素数是所给数字的因子则记录下来，更新所给数字 temp /= i。如果最后temp > 1，则也要将其输出。

Code：

#include <bits/stdc++.h>

using namespace std;

int main() {
    long long n;
    cin >> n;
    if (n == 1) {
        cout << n << "=" << n;
        return 0;
    } else {
        cout << n << "=";
    }
    vector<int> prime(50005, 1);
    prime[0] = prime[1] = 0;
    for (int i = 2; i * i < 50000; ++i) {
        for (int j = 2; i * j < 50000; ++j) {
            prime[i * j] = 0;
        }
    }
    bool isFirst = true;
    int temp = n;
    for (int i = 2; i < 50005; ++i) {
        int cnt = 0;
        bool found = false;
        while (prime[i] == 1 && temp % i == 0) {
            temp /= i;
            cnt++;
            found = true;
        }
        if (found) {
            if (isFirst) {
                cout << i;
                if (cnt > 1) cout << "^" << cnt;
                isFirst = false;
            } else {
                cout << "*" << i;
                if (cnt > 1) cout << "^" << cnt;
            }
        }
    }
    if (temp > 1) cout << (isFirst ? "" : "*") << temp << endl;
    return 0;
}