【C++】PAT(advanced level)1059. Prime Factors (25)

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

1.好吧， 乱写的，但是过了就不管了。

//#include<stdio.h>#include<iostream>#include<algorithm>#include<string>#include<string.h>#include<vector>#include<map>//#include<iomanip>using namespace std;int main(){//freopen("in.txt","r",stdin);long int m,mt;map <int,int> pr;map<int,int>::iterator it;scanf("%ld",&m);mt=m;bool ff=false;while(true){int p=2;while(true){if(p*p>m){ff=true;pr[m]++;break;}if(m%p==0){pr[p]++;m=m/p;break;}p++;}if(ff){break;}}cout<<mt<<"=";bool flag=true;for(it=pr.begin();it!=pr.end();it++){if(flag){flag=false;}else{cout<<"*";}cout<<(*it).first;if((*it).second!=1){cout<<"^"<<(*it).second;}}//fclose(stdin);system("pause");return 0;}