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Leftmost Digit(杭电1060)(求N^N的最高位)


Leftmost Digit


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13574    Accepted Submission(s): 5216


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


Hint


 



Author


Ignatius.L


/*本题思路: 
1,令M=N^N;

2,分别对等式两边取对数得 log10(M)=N*log10(N),得M=10^(N*log10(N));

3,令N*log10(N)=a+b,a为整数,b为小数;

4,C函数:log10(),计算对数,pow(a,b)计算a^b

5,由于10的任何整数次幂首位一定为1,所以,M的首位只和N*log10(N)的小数部分有关,

即只用求10^b救可以了;

6,最后对10^b取整,输出取整的这个数就行了。(因为0<=b<1,所以1<=10^b<10对

其取整,那么的到的就是一个个位,也就是所求的数)。
*/
//hdoj系统就是坑,各种CE,各种WA,多亏康晓辉指点。
#include<stdio.h>
#include<math.h>
int main()
{
int test;
double n,k,a;
scanf("%d",&test);
while(test--)
{
scanf("%lf",&n);
a=n*log10(n);
k=a-(__int64)a; //这里强制转换需要用int64才能A,否则会w或者CE!!!
printf("%d\n",int(pow(10,k)));
}
return 0;
}



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