HDU_1060_leftmostdigt

题目：

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

234

 

Sample Output

22
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.
 
//For each test case, you should output the leftmost digit of N^N.#include"stdio.h"#include"math.h"int comp(double num) {    double x = num * log10(num * 1.0);    double power = x - (long long)x;   <span style="color:#33ccff;"> </span><span style="color:#ff9900;">//32位的int对10^9会溢出，用64位整型long long or __int64。2^30 ≈ 10^9</span>    return (int)pow(10.0, power);}int main() {    int N;    scanf("%d", &N);    while(N--) {        double num;        scanf("%lf", &num);        int result = comp(num);        printf("%d\n", result);    }    return 0;}

技巧：取对数计算。

num ^ num = a * 10^n;
a * 10^n 为 num*num的科学计数法, 0 < a < 1, a 的整数部分恰为 num^num 的最左一位
log10(num * num) = log10(a * 10^n);
==> num* log10(num) = n + log10(a);
==> log10(a) = 0.******, 而n为正整数 ==> log10(a) 为 x = num * log10（num) 的小数部分;
==> x - n = log10(a) , a = 10^(x - n);
==> 所求为(int)a;

注意输入1,000,000,000时使用__int64, 或 long long 64位整型