HDOJ4278 Faulty Odometer[进制转化]

Faulty Odometer

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 222 Accepted Submission(s): 155

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

2012 ACM/ICPC Asia Regional Tianjin Online

Recommend

liuyiding

因为没有3和8，所以我们把一个数当作8（10-2）进制来看，因此，在这个输入的8进制数中，每一位的数上，9就看作是7，4、5、6、7

就当作3、4、5、6来看待，而3和8就当作0来看待，因此，当我们把这个八进制数转换成十进制数的时候，他就是我们所要求的结果了。

code:

 1 #include<iostream>
 2 using namespace std;
 3 
 4 int a[10]={0,1,2,0,3,4,5,6,0,7};
 5 
 6 int main()
 7 {
 8     int n,m;
 9     int ans;
10     int temp;
11     while(~scanf("%d",&n),n)
12     {
13         ans=0;
14         m=n;
15         temp=1;
16         while(n)
17         {
18             ans+=a[n%10]*temp;
19             n/=10;
20             temp*=8;
21         }
22         printf("%d: %d\n",m,ans);
23     }
24     return 0;
25 }