【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 *…*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

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;
}