Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1k1×p2k2×⋯×pmkm.
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
特判1=1
#include<bits/stdc++.h>
#pragma GCC optimize(3)
#define max(a,b) a>b?a:b
using namespace std;
typedef long long ll;
const int N=5e5+5;
int prime[N],cnt=0;
bool p[N];
void is_p(){
p[0]=p[1]=true;
for(int i=2;i<N;i++){
if(!p[i]) prime[cnt++]=i;
for(int j=0;j<cnt&&(ll)i*prime[j]<N;j++){
p[i*prime[j]]=true;
if(i%prime[j]==0) break;
}
}
}
struct node{
int p,k;
}a[N];
int main(){
is_p();
int n;
scanf("%d",&n);
int x=n;
printf("%d=",n);
if(n==1){
printf("%d\n",n);
return 0;
}
int t=0;
for(int i=0;i<cnt&&x!=1;i++){
if(x%prime[i]==0){
int res=0;
while(x%prime[i]==0){
res++;
x/=prime[i];
}
a[++t]={prime[i],res};
}
}
if(x!=1) a[++t]={x,1};
for(int i=1;i<=t;i++){
printf("%d",a[i].p);
if(a[i].k>1) printf("^%d",a[i].k);
printf("%c","*\n"[i==t]);
}
return 0;
}