zoukankan      html  css  js  c++  java
  • HDU4278

    Faulty Odometer

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 2102    Accepted Submission(s): 1456


    Problem Description

      You are given a car odometer which displays the miles traveled as an integer. The odometer has a defect, however: it proceeds from the digit 2 to the digit 4 and from the digit 7 to the digit 9, always skipping over the digit 3 and 8. This defect shows up in all positions (the one's, the ten's, the hundred's, etc.). For example, if the odometer displays 15229 and the car travels one mile, odometer reading changes to 15240 (instead of 15230).
     

    Input

      Each line of input contains a positive integer in the range 1..999999999 which represents an odometer reading. (Leading zeros will not appear in the input.) The end of input is indicated by a line containing a single 0. You may assume that no odometer reading will contain the digit 3 and 8.
     

    Output

      Each line of input will produce exactly one line of output, which will contain: the odometer reading from the input, a colon, one blank space, and the actual number of miles traveled by the car. 
     

    Sample Input

    15 2005 250 1500 999999 0
     

     

    Sample Output

    15: 12 2005: 1028 250: 160 1500: 768 999999: 262143
     

    Source

    八进制转十进制

     1 //2017-09-29
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <iostream>
     5 #include <algorithm>
     6 
     7 using namespace std;
     8 
     9 int digit[20], tot;
    10 
    11 int main()
    12 {
    13     int n;
    14     while(cin>>n && n){
    15         int ans = 0, tmp = n;
    16         tot = 0;
    17         while(tmp){
    18             int a = tmp%10;
    19             if(a >= 9)a -= 2;
    20             else if(a >= 4)a -= 1;
    21             digit[tot++] = a;
    22             tmp /= 10;
    23         }
    24         for(int i = tot-1; i >= 0; i--){
    25             ans *= 8;
    26             ans += digit[i];    
    27         }
    28         cout<<n<<": "<<ans<<endl;
    29     }
    30 
    31     return 0;
    32 }
  • 相关阅读:
    EL表达式学习
    IDEA maven项目创建速度慢
    c语言栈的链表实现
    c数据结构学习随笔
    自定义JSP中的Taglib标签之四自定义标签中的Function函数
    深入理解HTTP Session
    利用GeneratedKeyHolder获得新增数据主键值
    RBAC用户角色权限设计方案
    Fragment与Activity交互(使用接口)
    php session 自定义的设置测试
  • 原文地址:https://www.cnblogs.com/Penn000/p/7616736.html
Copyright © 2011-2022 走看看