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<iostream>#include<math.h>#include<map>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 << "^"<<it->second;            }        }else {            cout << "*" << it->first ;            if (it->second>1) {                cout << "^"<<it->second;            }        }    }    cout << endl;    return 0;}