Leftmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12912 Accepted Submission(s): 4943
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.
/* 题解：
看题目可知N的位数非常大，常规方法根本无法做出来，故这里用到一个
求N^N最高位a用到一个公式 a=10^(N*log10(N)-[N*log10(N)]) 注：[N*log10(N)]为向下取整
证明过程：
用科学计数法来表示 N^N = a*10^x;
比如 N = 3; 3^3 = 2.7 * 10^1;
我们要求的最右边的数字就是(int)a，即 a 的整数部分;
OK, 然后两边同时取以 10 为底的对数 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);
现在就只有 x 是未知的了，如果能用 n 来表示 x 的话，这题就解出来了。
又因为，x 是 N^N 的位数。比如 N^N = 1200 ==> x = 3;
实际上就是 x 就是 lg(N^N) 向下取整数，表示为[lg(N^N)]
a = 10^(N*lg(N) - [lg(N^N)]);
然后(int)a 就是答案
*/
#include<cstdio>
#include<cmath>
int main()
{
double m;
__int64 n;
int T;
scanf("%d",&T);
while(T--)
{
scanf("%I64d",&n);
m=n*log10(n+0.0);
m-=(__int64)m;
printf("%d\n",(int)pow(10,m));
}
return 0;
}
