题目1 : Binary Watch
时间限制:10000ms
单点时限:1000ms
内存限制:256MB
描述
Consider a binary watch with 5 binary digits to display hours (00 - 23) and 6 binary digits to display minutes (00 - 59).
For example 11:26 is displayed as 01011:011010.
Given a number x, output all times in human-readable format "hh:mm" when exactly x digits are 1.
输入
An integer x. (0 ≤ x ≤ 9)
输出
All times in increasing order.
样例输入
1
样例输出
00:01
00:02
00:04
00:08
00:16
00:32
01:00
02:00
04:00
08:00
16:00
EmacsNormalVim
题目大意
假设一个二进制手表用5个二进制指针表示 (00 - 23)小时用6个二进制指针表示 (00 - 59)分钟。
列如 11:26分显示成01011:011010。
给定一个数字x，输出所有人类可读的“hh:mm”格式的恰好x个指针为1的时间
代码
| #include <iostream> | |
| #include <cstring> | |
| #include <string> | |
| #include <set> | |
| #include <map> | |
| #include <fstream> | |
| #include <bitset> | |
| using namespace std; | |
| int main() | |
| { | |
| int x; | |
| cin>>x; | |
| for (int i = 0; i <24 ; ++i) { | |
| for (int j = 0; j < 60; ++j) { | |
| bitset<5> b1(i); | |
| bitset<6> b2(j); | |
| if( (b1.count()+b2.count())==x){ | |
| if(i<10){ | |
| cout<<"0"; | |
| } | |
| cout<<i<<":"; | |
| if(j<10){ | |
| cout<<"0"; | |
| } | |
| cout<<j<<endl; | |
| } | |
| } | |
| } | |
| return 0; | |
| } |
代码地址：
https://github.com/iptop/hihoCoderProblemSolving/blob/master/Binary%20Watch/main.cpp