HDU 1060 Leftmost Digit

题目：

Leftmost Digit

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

Total Submission(s): 3393 Accepted Submission(s): 1498

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.

思路：拿到这题，一看数据那么大，没有一点办法，想了很久，还是去看题解了。

令m=n^n

则logn（m）=n =>  lgm/lgn=n  =>  lgm=n*lgn  =>  m=10^(n*lgn);

10的整数次方得到肯定是1以开头的一个数，就要看10的小数次方得到那个数了，乘以小数次方得到的第一个数，就是我们求的数了

而且10的小数次方肯定是大于等于1，小于10的    因为10^0=1;  10^1=10;      

代码：

#include<stdio.h>
#include<math.h>
int main(){
int t;
scanf("%d",&t);
long long n;
while(t--){
scanf("%I64d",&n);
double x=(n*1.0)*log10(n*1.0);
x=x-(long long)x;
printf("%d\n",(int)pow(10.0,x));
}
return 0;
}