Leftmost Digit

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

Total Submission(s): 13880 Accepted Submission(s): 5304

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

234

Sample Output

22
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.
解题思路：
对于一个数num，用科学计数法可以表示为：num=10n + a ,使a的整数部分。即为num的最高位数字
本题举例：
         numnum=10n + a 这里的n和上述的n不等，只是均代表指数。
        两边取对数：num*lg(num) = n + lg(a) ；
                   由于a<10,所以0<lg(a)<1 ；
                 令x=n+lg(a) ；则n为x的整数部分，lg(a)为x的小数部分
                   又因为x=num*lg(num);即a=10(x-n)  =  10(x-int(x))
        再取a的整数部分即为num的最高位
源代码：
#include <stdio.h>#include <math.h>#include <stdlib.h>int main(){    __int64 cas,b,i,d;    double a,m,n,c;    scanf("%I64d",&cas);    for(i=1;i<=cas;i++)    {        scanf("%lf",&n);        a=n*log10(n);        b=(__int64)(a);        c=a-b;        d=(__int64)(pow(10,c));        printf("%I64d\n",d);    }    system("pause");        return 0;}