1059. Prime Factors (25)-PAT

1059. Prime Factors (25)

时间限制

50 ms

内存限制

32000 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

推荐指数：※

来源：http://pat.zju.edu.cn/contests/pat-a-practise/1059

这道题目主要是分解质因子。

1.考虑1的情况。

#include<iostream>#include<string.h>#include<math.h>using namespace std;bool is_prime(long n){long i,tmp;tmp=sqrt(n)+1;for(i=3;i<=tmp;i++){if(n%i==0)return false;}return true;}int main(){long n,i,tmp,k;cin>>n;cout<<n<<"=";if(n==1){cout<<1;return 0;}tmp=n;for(i=2;i<=tmp;i++){k=0;if(is_prime(i)==true){while(n%i==0){k++;n=n/i;}}if(k>1){cout<<i<<"^"<<k;}else if(k==1){cout<<i;}if(n!=1&k>=1)cout<<"*";else if(n==1)break;}return 0;}