PAT甲级真题 1059 Prime Factors (25分) C++实现（建立素数备忘录）

题目

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

思路

为了避免重复运算，维护一个素数表prime，初始化为{2}。

分解N时，先考察素数表中的数是否是其因子；若素数表中没有其因子，且素数表中最大值比根N小，那么逐个考察到根N，每发现一个素数都保存到素数表中。

把N的因子m保存到factor数组中，再迭代考察n / m；若其没有因子，则其本身就是素数，将其保存到factor中。

最后按指定规则输出结果即可。

注意输入的值可能是1，需单独考虑，否则测试点3无法通过。

代码

#include 
#include 
#include 
using namespace std;

vector<int> prime;  //素数备忘录
vector<int> factor;  //素因数
bool isPrime(int n){
    for (int i=2; i<=sqrt(n); i++){
        if (n % i == 0){
            return false;
        }
    }
    return true;
}
void divide(int n){
    if (n < 2){
        return;
    }
    int len = prime.size();
    int m = 0;
    for (int i=0; i<len; i++){
        if (n % prime[i] == 0) {
            m = n / prime[i];
            factor.push_back(prime[i]);
            break;
        }
    }
    if (m==0){  //n不能被备忘录中的素数整除，继续考察根N范围内是否有因数，扩展素数备忘录
        for (int i=prime[len-1]+1; i<=sqrt(n); i++){
            if (isPrime(i)){
                prime.push_back(i);
                if (n % i == 0){
                    m = n / i;
                    factor.push_back(i);
                    break;
                }
            }
        }
    }
    if (m==0){  //n确定是素数
        factor.push_back(n);
        return;
    }
    else {
        divide(m);
    }
}
int main(){
    prime.push_back(2);
    int n;
    cin >> n;
    cout << n << "=";
    if (n==1){
        cout << "1";
    }
    else{
        divide(n);
        int i = 0;
        while (i<factor.size()){
            if (i > 0){
                cout << "*";
            }
            cout << factor[i];
            int j = i + 1;
            while (j<factor.size() && factor[j]==factor[i]){
                j++;
            }
            int k = j - i;
            if (k > 1){
                cout << "^" << k;
            }
            i = j;
        }
    }
    return 0;
}