1059 Prime Factors (25)（25 分)

1059 Prime Factors (25)（25 分)

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

素数的应用，用的是厄拉多塞素数筛选法， 效率比较高。可见 https://blog.csdn.net/qq_34649947/article/details/79649247

 

#include 
using namespace std;
const int maxn = 100000 + 10;
int is_prime[maxn + 10];
int pNum = 0;
vector prime;
void get_prime() {
    for (int i = 2; i <= maxn; i++) {
        is_prime[i] = 1;
    }
    for (int i = 2; i*i <= maxn; i++) {
        if (is_prime[i]) {
            for (int j = i*i; j <= maxn; j += i) {
                is_prime[j] = 0;
            }
        }
    }
    for (int i = 2; i <= maxn; i++) {
    	if (is_prime[i]) {
    		pNum++;//在最大范围内所有的质数的数量
    		prime.push_back(i);//存储所有的质因子
    	}
    }
}
//题目要求有底数和指数，所有用一个结构体来存
struct factor {
	int x;//质因子
	int cnt;//个数
}fac[100];
int n;
int main() {
	get_prime();
	int n, num = 0;//num为不同质因子的个数
	cin >> n;
	if (n == 1) cout << "1=1" << endl;
	else {
		cout << n << "=";
		int sqr = (int)(sqrt(1.0*n));
		//枚举
		for (int i = 0; i < pNum && prime[i] <= sqr; i++) {
			if (n % prime[i] == 0) {
				fac[num].x = prime[i];
				fac[num].cnt = 0;
				while (n % prime[i] == 0) {
					fac[num].cnt++;
					n /= prime[i];
				}
				num++;
			}
			if (n == 1) break;//退出循环
		}
		//注意点，最后一个样例
		if (n != 1) {//如果无法被根号n以内的质因子除尽,那就是质因子本身;
			fac[num].x = n;
			fac[num].cnt = 1;//
			num++;
		}
		for (int i = 0; i < num; i++) {
			if (i > 0) cout << "*";
			cout << fac[i].x;
			if (fac[i].cnt > 1) {
				cout << "^" << fac[i].cnt;
			}
		}
	}
	return 0;

}