Faulty Odometer HDU - 4278 进制问题

Faulty Odometer

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2087 Accepted Submission(s): 1444

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

题意：行车记录仪记录的数会跳过3 跳过8

问行车记录仪实际应该记录的步数

tho: 尝试挺长时间才看出判断每个位置的数 然后加起来就行

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
long long int ave[12];
long long int f(long long int n)
{
int t;
int sum=0;
long long ans=0;
//    if( n%10 > 3 && n%10 <8) ans= n%10- 1;
//    else if(n%10>8) ans = n%10- 2;//最开始写的时候只想起了判个位数，
//    else ans = n%10;//后来改成分开判，最后发现合一块就行。。。
while(n)
{
t=n%10;
if(t>3&&t<8) t-=1;
if(t>8) t-=2;
ans += t*ave[sum];
sum++;
n/=10;
}
return ans;
}
int main()
{
int i,j;
long long int n;
int t=1;
ave[0]=0;
for(i=0;i<11;i++)
{
ave[i]=t;
t*=8;
}
while(cin>>n,n!=0)
{
long long ans=f(n);
printf("%d: %d\n",n,ans);
}
return 0;
}