【PAT】【Advanced Level】1059. Prime Factors (25)

1059. Prime Factors (25)

时间限制

100 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 *…*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 = p₁^k₁ * p₂^k₂ *…*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

原题链接：

https://www.patest.cn/contests/pat-a-practise/1059

https://www.nowcoder.com/pat/5/problem/4112

思路：

对于每一个数进行扫描，从2开始，每找到一个因子，不断相除，累加该因子次数。

——PAT AC，牛客网超时

对于每个数先判断是否是质数，如果是则不用再分解

——AC

CODE：

#include<iostream>#include<cstdio>#include<vector>using namespace std;typedef struct S{int nu;int p;};vector<S> pri;bool pr(int n){if (n==1||n==2||n==3) return 1;for (int i=2;i*i<=n;i++) if (n%i==0) return 0;return 1;}int main(){int n;cin>>n;cout<<n<<"=";if (pr(n)){ cout<<n; }else{int i=2;while (i<=n){if (n%i==0){S t;t.nu=i;t.p=0;while (n%i==0){t.p++;n/=i;}pri.push_back(t);}i++;}for (int i=0;i<pri.size()-1;i++){cout<<pri[i].nu;if (pri[i].p!=1){cout<<"^"<<pri[i].p; }cout<<"*";}cout<<pri[pri.size()-1].nu;if (pri[pri.size()-1].p!=1){cout<<"^"<<pri[pri.size()-1].p; }}return 0;}