hdu 1060 Leftmost Digit

Leftmost Digit

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

Total Submission(s): 16211 Accepted Submission(s): 6373

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

参考了好多别人的代码思路，终于理清了思路，原来math头文件里还有log()函数，现在是知道了，别人写的思路很清晰。

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；（这里挺关键的）

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

代码写的很简单，主要还是这种题的思路：

#include<stdio.h>
#include<math.h>
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
double n,a,b;
scanf("%lf",&n);
a = n*log10(n);
b = a - floor(a);  //得到小数部分
printf("%d\n",(int) pow(10,k));
}
return 0;
}