HDU 2.1.7 Leftmost Digit

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 <div style="" font-size:="" 14px;="" border-top:="" #b7cbff="" 1px="" dashed;="" font-family:="" times"=""> 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.

Author Ignatius.L

#include<stdio.h>  
#include"algorithm"  
using namespace std;  
int main()  
{  
    int testCase;  
    unsigned long long inNum;  
    scanf("%d", &testCase);  
    while (testCase--)  
    {  
        scanf("%lld", &inNum);  
         long double x = inNum*log10(inNum*1.0);  
         x -= unsigned long long(x);  
        printf("%d\n", int(pow(10.0, double(x))));  
    }  
    return 0;  
}  

【思路】

实数可以用科学记数法表示为： a=b*10^c，其中a是任意实数，1<=b<10，C为整数。

此题中，设输入为m,那么 m^m=k*10^n，本题要求解的就是k的整数部分，也就是int(k)

两边取对数：m*log10(m)=n+log10(k)

设x=m*log10(m)=n+log10(k)，

由于1<=k<10,所以0<=log10(k)<1

n是x的整数部分。log10(k)是x的小数部分。

换句话说，log10(k)=[x的小数部分]=[x-int(x)]=[m*log10(m)-int(m*log10(m))]

所以int(k)=int(10^log10(k))=int(10^[m*log10(m)-int(m*log10(m))])

【注意事项】

由于m比较大，所以要用到unsign long long,注意输入时的%lld不要出错。