hdu acmsteps 2.1.8 Leftmost Digit

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
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<cmath>
using namespace std;
int main(){
    int t;
    unsigned long long n;
    cin>>t;
    while(t--){
        cin>>n;
        long double x=n*log10(n*1.0);
        x-=unsigned long long int(x);
        int ans=pow(10.0,double(x));
        cout<<ans<<'
';
    }
}

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
       number^number=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.