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 = 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
很简单的一个小技巧,从i=2开始向上遍历,当遇到 N % i == 0时,把N一直向下除,直到余i不等于0位置,那么我们可以得到每个N%i==0的i都会是素数,这里和欧拉函数的计算十分类似
但是特别要注意当N == 1时,要特殊判断
#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;
}