hdu 4278 Faulty Odometer

最新推荐文章于 2021-08-23 10:23:03 发布

转载最新推荐文章于 2021-08-23 10:23:03 发布 · 88 阅读

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原文链接：http://www.cnblogs.com/8023spz/p/9002469.html

本文介绍了一种特殊的里程表转换算法，该里程表跳过了数字3和8，导致每进一位只使用8个数字而不是通常的10个。文章详细解释了如何将这种特殊里程表显示的数值转换为实际行驶的里程，并附带了完整的C++实现代码。

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

Sample Output

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

本来是十进制，但是少了3和8，每个位现在都用八个数来计数，4代表的实际是3,5-4,6-5,7-6,9-7，其实就是没以为加够了八个数就会进位，当成八进制了，先把每一位变成他该表示值，那么这个数整个就变成了八进制的表示，转化成十进制就ok。
代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
char s[10],t[10];

int main()
{
    while(cin>>s && strcmp(s,"0"))
    {
        int i = -1;
        strcpy(t,s);
        while(s[++ i])
        {
            if(s[i] > '8')s[i] -= 2;
            else if(s[i] > '3')s[i] -= 1;
        }
        cout<<t<<": ";
        int sum = 0,d = 1;
        while(-- i >= 0)
        {
            sum += (s[i] - '0') * d;
            d *= 8;
        }
        cout<<sum<<endl;
    }
}