题解 | #广告屏幕#
广告屏幕
https://ac.nowcoder.com/acm/problem/22213
using namespace std;
#define int long long
#define endl '\n'
signed main(){
std::ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int n;cin>>n;int a[n],b[n],i=0,j=0;
for(int x=1;x<=n;x++){
for(int y=n/x;y>=x;y--){//从n/x开始找，快一些，要是从n开始会运行超时
if(x*y==n)
{
a[i]=x,b[j]=y;
i++,j++;
}
}
}
cout<<a[i-1]<<" "<<b[j-1]<<endl;
return 0;
}
