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 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. 题意：给你一个数n让你求n^n最高位的数； 解题思路：那天上午在实验室写了这个题，刚开始看到数很大就寻思找规律，结果输出不到20位，数就太大了，回来之后想了好几天，后来又借鉴了大神的思路，知道了。 num^num=10^n*a;a的整数部分就是这个数的最左边的那一位； 对两边取对数  num*log10（num）= n+lg(a);n+lg(a)这个数中，n肯定是个整数，lg（a）肯定是个小数； 代码： #include using namespace std; int main() {     //freopen("in.txt","r",stdin);     unsigned long n;     int t;     scanf("%d",&t);     while(t--)     {         scanf("%ld",&n);         double x=n*log10(n*1.0);         //cout<<"(__int64)x="<<(__int64)x<<endl;         x-=(__int64)x;         //cout<<"x="<<x<<endl;         int a=pow(10.0, x);         printf("%d\n",a);     }     return 0; }