hdu1060

N^N=10^k

N*lg(N)=k;

p=k-(int)k;

答案是10^p

 

求N^N的最左的一位数字

#include <iostream>#include <cmath>#include <cstdio>using namespace std;int main(){    int T;    int n;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        double p=n*log10((double)n);        p=p-(long long int)p;        printf("%d\n",(int)pow(10.0,p));    }    return 0;}