其实不用判断是不是质数的。题目说N是两个质数的乘积,所以直接暴力枚举因数就好了。
by Jellychen @ 2024-02-08 11:45:19
先找质数然后暴力
```cpp
#include<iostream>
#include<bitset>
#include<algorithm>
#include<vector>
using namespace std;
auto getPrimeList()
{
const unsigned int maxNum = 1000000;
bitset<maxNum> compositeTagList;
vector<unsigned int> primeList = {};
for (unsigned int i = 2; i < maxNum; i += 1)
{
if (compositeTagList[i] == false)
{
for (unsigned long int j = i * 2; j < maxNum; j += i)
{
if (compositeTagList[j] == false)
compositeTagList[j] = true;
}
primeList.push_back(i);
}
}
return primeList;
}
auto primeList = getPrimeList();
int main()
{
unsigned int n;
cin >> n;
for (const unsigned int &x: primeList) {
if (n % x == 0)
{
cout << n / x;
break;
}
}
return 0;
}
```
by 145a @ 2024-03-01 09:04:01
你知道什么叫平方根吧？这题可以用上
by 13860121259AaBb @ 2024-03-03 20:15:51