突破口:nn=10nlog10(n),利用科学计数法,nn=10k×10m,0<=k<1,k+m=nlog10(n)。
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
int T,i;
long long N,ans;
double n;
scanf("%d",&T);
while(T--)
{
scanf("%lld",&N);
n=(double)N;
n=n*log10(n);//printf("%lf\n",n);
n=n-(long long)n;//printf("%lf\n",n);
ans=(long long)pow(10,n);
printf("%lld\n",ans);
}
return 0;
}