PAT甲级1059 Prime Factors (25 分)

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

结尾无空行

Sample Output:

97532468=2^2*11*17*101*1291

结尾无空行

/*思路：素数打表*/
#include <bits/stdc++.h>
using namespace std;
const int maxv=100010;
int prime[maxv],num=0;
struct numx
{
    int x,cnt;
};
bool judge(int n)
{
    if(n<2) return false;
    for(int i=2;i*i<=n;i++) if(n%i==0) return false;
    return true;
}
void find_prime()
{
    for(int i=2;i<maxv;i++)
    {
        if(judge(i)) prime[num++]=i;
    }
}
int main()
{
    int n;
    vector<numx> v;
    scanf("%d",&n);
    find_prime();
    if(n==1) printf("1=1\n");
    else
    {
        printf("%d=",n);
        int sqr=(int)sqrt(n*1.0);
        for(int i=0;prime[i]<=sqr;i++)
        {
            if(n%prime[i]==0)
            {
                numx temp;
                temp.x=prime[i];
                temp.cnt=0;
                while(n%prime[i]==0)
                {
                    temp.cnt++;
                    n/=prime[i];
                }
                v.push_back(temp);
                if(n==1) break;
            }
        }
    }
    if(n!=1) v.push_back({n,1});
    for(int i=0;i<v.size();i++)
    {
        if(v[i].cnt>1) printf("%d^%d",v[i].x,v[i].cnt);
        else printf("%d",v[i].x);
        if(i!=v.size()-1) printf("*");
    }
    return 0;
}