HDU1060：Leftmost Digit（数论）

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 17765    Accepted Submission(s): 6869


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.

 

Author

Ignatius.L
题意：求N^N的最高位数字。

思路：N^N用科学记数法表示，N^N = (double)a * 10^x，x为N^N的位数减一，用N去表示x显然有x = lg(N^N)向下取整；所以(double)a = N^N / 10^( (int)lg(N^N) )，a的整数部分就是答案，但直接计算会溢出，两边取对数有lg(a) = N*lg(N) - (int)N*lg(N)，a = 10^( N*lg(N) -
(int)N*lg(N) )，这样就不会溢出了。

# include <stdio.h>
# include <math.h>
int main()
{
int t, n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
double k = n*log10(n);
k -= (long long)k;
printf("%lld\n",(long long)pow(10, k));
}
return 0;
}