Leftmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10997 Accepted Submission(s): 4209
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.
分析：
m=n^n,两边分别对10取对数得 log10(m)=n*log10(n),
得m=10^(n*log10(n)),由于10的任何整数次幂首位一定为1,
所以m的首位只和n*log10(n)的小数部分有关
得m=10^(n*log10(n)),由于10的任何整数次幂首位一定为1,
所以m的首位只和n*log10(n)的小数部分有关
#include<stdio.h>
#include<math.h>
int main()
{
int n,t;
double a,a1,a2;
scanf("%d", &t);
while (t--)
{
scanf("%d",&n);
a=n*log10(n*1.0);
a1=a-(__int64)a;
a2=pow(10.0,a1);
printf("%d ", (int)a2);
}
return 0;
}
#include<math.h>
int main()
{
int n,t;
double a,a1,a2;
scanf("%d", &t);
while (t--)
{
scanf("%d",&n);
a=n*log10(n*1.0);
a1=a-(__int64)a;
a2=pow(10.0,a1);
printf("%d ", (int)a2);
}
return 0;
}