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Leftmost Digit（经典数学问题，对数求解）

2016-07-21 19:44 387 查看

Leftmost Digit

 
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3372 Accepted Submission(s): 1482
[align=left]Problem Description[/align]
Given a positive integer N, you should output the leftmost digit of N^N.

 
[align=left]Input[/align]
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).

 
[align=left]Output[/align]
For each test case, you should output the leftmost digit of N^N.

 
[align=left]Sample Input[/align]

2
3
4


 用科学计数法推规律：
#include <iostream>
#include <math.h>
using namespace std;
int main()
{
int t;
while(cin>>t)
{
while(t--)
{
unsigned long n;
cin>>n;
double x=n*log10(n*1.0);
x-=(__int64)x;
int a=pow(10.0, x);
cout<<a<<endl;
}
}
return 0;
}


[align=left]Sample Output[/align]

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.

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