Leftmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4901 Accepted Submission(s): 1793
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.
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std
;
int main()
{
__int64 ci
;scanf
("%I64d",&ci
);
while(ci
--)
{
__int64 n
;scanf
("%I64d",&n
);
double l
=n
*log10
(n
+0.0);
__int64 m
=(__int64)pow
(10.0,l
-(__int64)l
);
printf
("%I64d/n",m
);
}
return 0;
}
