杭电acm hdu-1060Leftmost Digit

                     Leftmost Digit解答

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12972    Accepted Submission(s): 4966

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

1. 设n的n次方==x*10的a次方则
2. n*㏒n=㏒x+a      a是n㏒n的整数部分，㏒x是他的小数部分
3. ㏒x=n*㏒n-a      且a=（int）n㏒n
4. ㏒x=n*㏒n-（int）n*㏒n下面就好懂了
5. 代码如下
6.  
7.  

8. #include<stdio.h>#include<math.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);    }    return 0;}