hdu 1060 Leftmost Digit

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13503    Accepted Submission(s): 5187

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.
题目大意：求n的n次方的最左边的那个数思路：数学公式，一个数的位数等于(int)lg(n)+1 ;2014,11,7 感觉还是有点毛病
#include<stdio.h>#include<math.h>int main(){int t,n,b;__int64 m;scanf("%d",&t);while(t--){scanf("%d",&n);m=(__int64)(n*log10(n));//n的n次方的位数是n*lg(n)+1,又因为A=b*10m次方，所以//m等于位数-1，即n*lg(10);b=pow(10.0,(n*log10(n)-m));//lgb=n*lg(n)-m printf("%d\n",b);}return 0;}