Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11600 Accepted Submission(s): 4430
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
一个数n^n=p= x*10^m 所以log(p)=n*log10(n);
而这和我们要求的最高位i的那个数有什么关系乐 比如求1234的最高位为1,而log10(1234)=log10(1.234*10^3)=log10(1.234)+3;
而最高位即为其pow(10,1og10(1.234))=1.234的整数1;
=log10(n^n)=n*log10(n);所以同样的道理,就可以求这个了.....
代码:
1 #include<stdio.h>
2 #include<math.h>
3 int main()
4 {
5 int n,m;
6 scanf("%d",&m);
7 while(m--)
8 {
9 scanf("%d",&n);
10 double a=n*log10(1.0*n);
11 a-=(__int64)a;
12 printf("%d\n",(int)pow(10,a));
13 }
14 return 0;
15 }