HDU 1060 - Leftmost Digit

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

Total Submission(s): 12528    Accepted Submission(s): 4784


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

 

Recommend

We have carefully selected several similar problems for you:  1018 1061 1071 1573 1066 

 

这题让我认识到了数学真是太伟大了！！！！
几十行不过和13行秒过的区别！！
这题写的时候有改动，没有加循环输入，直接submit HDU会WA

#include<stdio.h>
#include<string.h>
#include<math.h>

int main(){
int n;
while(scanf("%d",&n)!=EOF){		//以10为底求对数，得到总位数
double tmp=n*log10(n);		//tmp=X.abcdefghijk，10^X为最高位权值，0.abcdefghijk为最高位基数
double res=tmp-floor(tmp);	//floor向下取整函数
printf("%d\n",(int)pow(10,res));
}
return 0;
}