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).
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 思路:题目要求的是n^n的最左边的数字,因为n的范围很大,直接求是不可能的,那么就要用一下数学的思想来转化。假设我们要求的是1234的最左边的数字,我们可以先转化为求1.234的整数位,怎么转化呢?因为log10(1234)-log(1000)=log10(1234。0/1000)=log10(1.234),呵呵,发现问题了吗?其中log10(1000)=(int)log(1234)。那么就有log10(1.234)=log10(1234)-(int)log(1234),最后就只需要求pow(10,log10(1.234))的整数位就好了。把n^n换回去又有log10(n^n)=nlog10(n)。#include <stdio.h> #include <math.h> int main() { int n,x; scanf("%d",&n); while(n--) { scanf("%d",&x); printf("%d\n",(int)pow(10,x*log10(x)-(__int64)(x*log10(x))));//x*log10(x) int可能会存不下 } return 0; }