Leftmost Digit

Problem Description Given a positive integer N, you should output the leftmost digit of N^N.

Input The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow. Each test case contains a single positive integer N(1<=N<=1,000,000,000).

Output For each test case, you should output the leftmost digit of N^N.

Sample Input 2 3 4

Sample Output 2 2 Hint In the first case, 3 * 3 * 3 = 27, so the leftmost digit is 2. In the second case, 4 * 4 * 4 * 4 = 256, so the leftmost digit is 2. 先将N^N对数化，取10为底，又N^N可表示为a*10^x,其中9>=a>=1,故x为N^N的位数-1，eg：100=1*10^2 故x为2 lg(N^N) = lg(a*10^x) ; 化简        N*lg(N)  = lg(a) + x; 继续化       N*lg(N) - x = lg(a)             a = 10^(N*lg(N) - x); 问题转化成求x： 因为1<=a<=9，所以lg（a）<1; 故x=（int） N*lg(N) 向下取整 最后（int）a就是最后的结果   View Code 1 #include<iostream> 2 #include<cmath> 3 #include<cstdio> 4 using namespace std; 5 int main() 6 { 7 int n; 8 __int64 m; 9 cin>>n; 10 while(n--) 11 { 12 scanf("%I64d",&m); 13 double t = m*log10(m*1.0); 14 t -= (__int64)t; 15 __int64 ans = pow((double)10, t); 16 printf("%I64d\n",ans); 17 } 18 return 0; 19 }