首页 > 代码库 > HDU--1060
HDU--1060
Leftmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14696 Accepted Submission(s): 5660
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).
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 2HintIn 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
description:求解N^N的最左边的数字。
解法:num=N^N,两边取对数可得num=10^(N*log10(N))。当中10的整数次方的最左边总是数字10的整数倍。而决定num最左边的因素为N*log10N的小数部分。由此可解!
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> using namespace std; int main() { int t; cin >> t; while (t--) { long long int num; cin >> num; double num1 = num * log10( double (num)); long long int num2 = (long long int) num1; double num3 = num1 - num2; num = (long long int) pow(10,num3); cout << num<< endl; } return 0; }
做不出本题数学是硬伤啊!!。。。。
HDU--1060
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。