Leftmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 19366 Accepted Submission(s): 7677 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. 设N^N=x ,两边取对数:log(N*N)=N*log N=log x,10^(N*log N)=x,因为10的任何整数次方的首位都是1,所以x的首位数和N*log N的小数部分有关
#include#include #include #include #include #define ll long longusing namespace std;int main(){ int t; ll n; double x; cin>>t; while(t--) { cin>>n; x=n*log10((double)n); x-=(ll)x; x=(int)pow(10,x); cout< <