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:
    hust 1605 bfs
    hdu 1512
    2013 ACMICPC 杭州现场赛 I题
    2013年 ACMICPC 杭州赛区H题
    hdu 3717 二分+队列维护
    hdu 2993 斜率dp
    hdu 3480 斜率dp
    hdu 3507 斜率dp
    hdu 2829 斜率DP
    零碎笔记
  • 原文地址:https://www.cnblogs.com/Penn000/p/7616736.html
Copyright © 2011-2022 走看看