poj-2719【进制转换】

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Faulty Odometer

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9289 Accepted: 5751

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 3 to the digit 5, always skipping over the digit 4. This defect shows up in all positions (the one's, the ten's, the hundred's, etc.). For example, if the odometer displays 15339 and the car travels one mile, odometer reading changes to 15350 (instead of 15340).

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 4.

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

131520032005239250139915009999990

Sample Output

13: 1215: 132003: 14612005: 1462239: 197250: 1981399: 10521500: 1053999999: 531440

题意：有一种很奇怪的计数法，它的数里面没有4，现在用这种计数法给定一个数字n，把它换算成通用的十进制。

思路：正确的思路是采用9进制。把各个位上的大于4的数都减一， 例如：15变成14，67变成51，然后把这个数字当成是9进制。

#include<cstdio>#include<cstring>using namespace std;char str[20];int main(){while(scanf("%s",str),str[0]!='0'){int len=strlen(str);int ans=str[len-1]-'0'-(str[len-1]>'4');int bit=9;for(int i=len-2;i>=0;i--){if(str[i]>'4')ans+=(str[i]-'0'-1)*bit;elseans+=(str[i]-'0')*bit;bit*=9;}printf("%s: %d\n",str,ans);}return 0;}