hdoj1060Leftmost Digit

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15822    Accepted Submission(s): 6172

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.
 
Author

Ignatius.L

math函数的应用。。。log。。。

代码：

#include<cstdio>#include<cstring>#include<cmath>int main(){    int t;scanf("%d",&t);    while (t--)    {        long long p,n;        double m;         scanf("%lld",&n);        m=n*log10(n*1.0);        //printf("%lf\n",m);        p=m;        m=m-p;        printf("%d\n",(int)pow(10.0,m));    }    return 0;}