hdu1060 N^N首位 log10

一个数a必然可以表示为10^(c+d),其中c为整数，d为小数。则n^n的首位取决于d的第一个有效数字。10^d是讲小数点后移的。

#include <iostream>#include <string>#include <cstring>#include <queue>#include <stack>#include <cmath>#include <algorithm>#include <cstdio>using namespace std;int main(){    int t;    cin >> t;    while(t -- )    {        double sum , per , beh , n;        cin >> n;        beh = n * log10(n) - (long long)(n* log10(n));        per = pow((double)10 , beh);        per = (long long )per;        cout << per << endl;    }    return 0 ;}