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Given any positive Integer N,you are supposed to find all of its prime factors,and write them in the format N = @H_262_37@p?1???k?1????×@H_262_37@p?2???k?2????×?×@H_262_37@p?@H_375_53@m???k?@H_375_53@m????.
Each input file contains one test case which gives a positive Integer N in the range of long int.
Factor N in the format N = p?1??^k?1??*p?2??^k?2??*…*p?@H_375_53@m??^k?@H_375_53@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.
97532468
97532468=2^2*11*17*101*1291
/* Name: Copyright: Author: 流照君 Date: 2019/8/6 11:09:58 Description: */ #include <iostream> #include<String> #include <algorithm> #include <vector> #include<cmath> #define inf 0x3f3f3f using namespace std; typedef long long ll; ll prime[inf],a[inf]; ll num=1; void sieve(int n) { for(int i=2;i<=n;i++) { if(a[i]==0) prime[num++]=i; for(int j=i*2;j<=n;j=j+i) { a[j]=1; } } } int main(int argc,char** argv) { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); fill(a,a+inf,0); ll n,flag=0; cin>>n; sieve(500000); //for(int i=1;i<=num;i++) //cout<<prime[i]<<" "; //cout<<endl; if(n==1) { cout<<n<<"="<<"1"; return 0; } cout<<n<<"="; for(int i=1;i<num;i++) { int sum=0; while(n%prime[i]==0) { n=n/prime[i]; sum++; } if(flag&&sum) { if(sum==1) cout<<"*"<<prime[i]; if(sum>=2) { cout<<"*"<<prime[i]<<"^"<<sum; } } if(flag==0&&sum) { if(sum==1) cout<<prime[i]; if(sum>=2) { cout<<prime[i]<<"^"<<sum; } flag=1; } if(n==1) return 0; } return 0; }
别忘了考虑特例 1
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