HDU4278 Faulty Odometer

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Description

　　You are given a car odometer(里程表) which displays the miles traveled as an integer(整数). The odometer has a defect(缺点), however: it proceeds(收入) from the digit(数字) 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(投入) contains a positive(积极的) 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(表明) by a line containing a single 0. You may assume(承担)that no odometer reading will contain the digit 3 and 8.

Output

　　Each line of input will produce exactly one line of output(输出), 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

15200525015009999990

Sample Output

15: 122005: 1028250: 1601500: 768999999: 262143

这题我做完之后上网搜了一下，发现都是说是由八进制转化为十进制，想了一下，确实是这么回事，然而对我这个各种进制全凭自学的人来说，想转化成代码可不容易，而我自己的想法，就是利用dp找每一位所能包含的不含3,8的数的个数，然后一位位加过去就行。

#include <stdio.h>#include <cstring>int dp[15],v[15],l;void init(){int i;dp[0]=1;for(i=1;i<10;i++){dp[i]=dp[i-1]*8;}}int main(){int i,j;int n;init();while(~scanf("%d",&n)&&n){int ans=0;l=0;int no=n;while(n){v[l++]=n%10;n=n/10;}for(i=l-1;i>=0;i--){if(v[i]>3&&v[i]<8)v[i]--;else if(v[i]>8)v[i]=v[i]-2;ans+=dp[i]*v[i];}printf("%d: %d\n",no,ans);}return 0;}

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