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 = p​1​​​k​1​​​​×p​2​​​k​2​​​​×⋯×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 = p​1​​^k​1​​*p​2​​^k​2​​*…*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

Sample Input:

97532468

Sample Output:

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

很簡單的一個小技巧，從i=2開始向上遍歷，當遇到 N % i == 0時，把N一直向下除，直到餘i不等於0位置，那麼我們可以得到每個N%i==0的i都會是素數，這裡和尤拉函式的計算十分類似

但是特別要注意當N == 1時，要特殊判斷

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <cmath>
#define INF 0x3f3f3f3f

using namespace std;
typedef long long ll;
const int maxn = 1e6;
ll n;
int prime[maxn],num[maxn];
int main()
{
    int index = 0;
    scanf("%lld",&n);
    if(n == 1)
        printf("1=1\n");
    else
    {
        ll ans = n;
        for(ll i = 2;;i ++)
        {
            if(ans == 1) break ;
            if(ans % i == 0){
                prime[index++] = i;
                while(ans % i == 0)
                {
                    ans /= i;num[index-1]++;
                }
            }
        }
        printf("%lld=",n);
        for(int i = 0;i < index;i ++)
            if(num[i] == 1)
                printf("%d%c",prime[i],i==index-1?'\n':'*');
        else
            printf("%d^%d%c",prime[i],num[i],i==index-1?'\n':'*');
    }

    return 0;
}