HDU1060 数论

注意会发生溢出，用long long

思想：转换成N*log10(N)

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define INF 0x3f3f3f
int main()
{		
	int T;
	cin >> T;
	while(T--)
	{	
		ll n;
		cin >> n;
		double t1=(double)n*(log10((double)n));
		double t2=t1-(ll)t1;
		cout << (floor)(pow(10.0,t2))<< endl;
	}
	return 0;
}