Hdu 4278 Faulty Odometer(8进制转10进制)

Faulty Odometer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1664    Accepted Submission(s): 1146

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

 

题意：从1数到10，会跳过3和8。给你一个数，求这个数原来的值。

题解：4被当成3用，5 6 7则代表4,5,6，9代表8，所以这类似8进制数。数值处理下转换成10进制就是原来的值。

#include<cstring>
#include<algorithm>
#include<iostream>
#include<cstdio>
#define N 5005
#define ll long long

using namespace std;

int a[]= {0,1,2,0,3,4,5,6,0,7,0};
ll k;

int main() {
    //freopen("test.in","r",stdin);
    while(~scanf("%I64d",&k)&&k) {
        int num[10];
        int l=0;
        ll kk=k;
        while(k) {
            num[l++]=k%10;
            k/=10;
        }
        ll p=1;
        ll ans=0;
        for(int i=0; i<l; i++) {
            ans+=a[num[i]]*p;
            p*=8;
        }
        printf("%I64d: %I64d
",kk,ans);
    }
    return 0;
}