H - Leftmost Digit
Given a positive integer N, you should output the leftmost digit of N^N.
InputThe 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).
OutputFor each test case, you should output the leftmost digit of N^N.
Sample Input
2 3 4Sample 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.思路:这个题目的话,可以说想法是很巧妙的,用的了 log 函数,加上了公式的推导,具体请看公式推导过程
首先用科学计数法来表示 N^N = a*10^x; 比如N = 3; 3^3 = 2.7 * 10^1;
我们要求的最右边的数字就是(int)a,即a的整数部分;
OK, 然后两边同时取以10为底的对数 lg(N^N) = lg(a*10^x) ;
化简 N*lg(N) = lg(a) + x;
继续化 N*lg(N) - x = lg(a)
a = 10^(N*lg(N) - x);
现在就只有x是未知的了,如果能用n来表示x的话,这题就解出来了。
又因为,x是N^N的位数。比如 N^N = 1200 ==> x = 3; 实际上就是 x 就是 lg(N^N) 向下取整数,
表
示为[lg(N^N)]好啦 a = 10^(N*lg(N) - [lg(N^N)]); 然后(int)a 就是答案了。
代码:
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
int main()
{
int n;
scanf("%d",&n);
while(n--)
{
int a;
scanf("%d",&a);
double m=(double)a*log10(a);
m-=(long long)m;
int t=(int)pow(10,m);
printf("%d\n",t);
}
return 0;
}