PAT 1059. Prime Factors (25)（分解质因数）

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1059. Prime Factors (25)

时间限制
50 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 *…*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

解题思路

• 1.题目要求将一个数分解质因数并打印出来。
• 2.从小到大求质因数，保存上一个质因数，这样求下一个质因数就可以从这个数开始算了，避免重复。

AC代码

#include
#include
#include
using namespace std;
map<int,int> p;
int isPrime(int a,int fomer){
    for (int i = fomer; i <= sqrt(a); ++i) {
        if (a % i == 0) {
            return i;
        }
    }
    return 0;
}

int main(int argc, char *argv[])
{
    int n;
    cin >> n;
    int tem_n = n;
    int fomer =2;
    while (1) {
    //记录上一个质因数
        fomer = isPrime(n,fomer);
        if (fomer == 0) {
            if (p.find(n)==p.end()) {
                p[n] = 1;
            }else {
                p[n]++;
            }
            break;
        }
        //把所有质因数保存在p中
        if (p.find(fomer)==p.end()) {
            p[fomer] = 1;
        }else {
            p[fomer]++;
        }
        n = n / fomer;
    }
    //打印
    map<int,int>::iterator it;
    cout << tem_n << "=";
    for(it = p.begin();it!=p.end();it++){
        if (it == p.begin()) {
            cout << it->first ;
            if (it->second>1) {
                cout << "^"<second;
            }
        }else {
            cout << "*" << it->first ;
            if (it->second>1) {
                cout << "^"<second;
            }
        }
    }
    cout << endl;
    return 0;
}