Leftmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1982 Accepted Submission(s): 884
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 2Hint
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.
1: #include<iostream>
2: #include<cmath>
3: using namespace std;
4: int main(){
5: int t;
6: unsigned long long n;
7: cin>>t;
8: while(t--){
9: cin>>n;
10: long double x=n*log10(n*1.0);
11: x-=unsigned long long int(x);
12: int ans=pow(10.0,double(x));
13: cout<<ans<<' ';
14: }
15: }
解释:
对一个数num可写为 number=10n * a, 即科学计数法,使a的整数部分即为num的最高位数字
numbernumber=10n * a 这里的n与上面的n不等
两边取对数: number*log10(number) = n + log10(a);
因为a<10,所以0<log10(a)<1
令x=n+log10(a); 则n为x的整数部分,log10(a)为x的小数部分
又x=number*log10(number);
a=10(x-n) = 10(x-int(x)))
再取a的整数部分即得number的最高位.