Leftmost Digit
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 7 Accepted Submission(s) : 2
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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).
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.
In the second case, 4 * 4 * 4 * 4 = 256, so the leftmost digit is 2.
题意：
给你一个数字n。输出n的n次的第一位。
题解：
n^n=n*log10(n)。而10的整数次方数的首位为1，因此要求此题的答案仅仅需求出n*log10(n)的小数部分m。由于0<m<10,所以10^m的整数部分（(int)pow(10,m)）即为答案。
參考代码：
#include<stdio.h>
#include<math.h>
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
double a,c,n;
long long d,b;
scanf("%lf",&n);
a=n*log10(n);
b=(long long)a;//因为题目数据量的限制。这里的b的强制转换为long long型。int型会wa。
c=a-b;
d=(int)pow(10.0,c);
printf("%lld
",d);
}
return 0;
}