1.题目要求求一个数的质因数分解。
2.一个大于等于2的数，质因数分解分两种情况：
1）如果这个数是质数，那么质因数分解就是它本身；
2）如果不是质数，那么除去最大的质因数后，剩下的质因数均小于sqrt（n），所以遍历到sqrt（n）即可。
3.注意输入为1的情况，输出应该为1=1。
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=2^2*11*17*101*1291
AC代码：
[c language=”++”]
//#include<string>
//#include<stack>
//#include<unordered_set>
//#include <sstream>
//#include "func.h"
//#include <list>
#include <iomanip>
#include<unordered_map>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include <algorithm>
#include<stdio.h>
#include<iostream>
#include<string>
#include<memory.h>
#include<limits.h>
#include<stack>
using namespace std;
int main(void)
{
int n;
cin >> n;
int outPur = n;
vector<pair<int, int>> factor(0);
int tmp = sqrt(n) + 1;
for (int i = 2; i <= tmp; i++)
{//遍历到sqrt（n）+1即可
if (n%i == 0)
{
factor.push_back({ i, 1 });
n /= i;
while (n%i == 0)
{
factor.back().second++;
n /= i;
}
}
}
if (n != 1)//如果最后不为1，那么这个是最大质因数
factor.push_back({ n, 1 });
if (outPur == 1)
{//注意1的情况
printf("1=1\n");
return 0;
}
printf("%d=", outPur);
for (int i = 0; i < factor.size(); i++)
{
printf("%d", factor[i].first);
if (factor[i].second != 1)
printf("^%d", factor[i].second);
if (i != factor.size() – 1)
printf("*");
}
cout << endl;
return 0;
}
[/c]