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=2^211171011291
先用素数晒法把求出素数然后再进行枚举
#include<bits/stdc++.h>
using namespace std;
#define ll long long
int a[100005];
int prime[100005];
int tot;
void make_prime()
{
memset(a,1,sizeof(a));
a[0]=0,a[1]=0;
tot=0;
for(int i=2; i<=100005; i++)
{
if(a[i])
{
prime[tot++]=i;
for(int j=i+i; j<=100005; j+=i)
{
a[j]=0;
}
}
}
}
int main()
{
make_prime();
ll n;
scanf("%lld",&n);
if(n==1)
{
printf("1=1");
return 0;
}
printf("%lld=",n);
int cnt=0;
for(int i=0; i<tot&&n>=2; i++)
{
cnt=0;
while(n%prime[i]==0)
{
n/=prime[i];
cnt++;
}
if(cnt==0) continue;
if(cnt==1)
{
printf("%d",prime[i]);
}
else
{
printf("%d^%d",prime[i],cnt);
}
if(n>1)
{
printf("*");
}
}
}
