【杭电】4278八进制转化为十进制_12345是八进制,用编程转为十进制-优快云博客

本文链接：https://blog.youkuaiyun.com/idealistic/article/details/62894885

本文介绍了一种特殊算法，用于解决具有缺陷的汽车里程表显示问题。该里程表从2跳到4，从7跳到9，忽略数字3和8。文章通过示例解释了如何将这种非标准计数方式转换为实际行驶里程，并提供了实现这一转换的C++代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Faulty Odometer

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2023 Accepted Submission(s): 1400

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

Sample Output

  
   15: 12
2005: 1028
250: 160
1500: 768
999999: 262143

Source

2012 ACM/ICPC Asia Regional Tianjin Online

Recommend

liuyiding | We have carefully selected several similar problems for you: 4267 4268 4269 4270 4271

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
	char s[100];
	int num[100];
	while(~scanf("%s",s)&&s[0]!='0'){
		int l=strlen(s);
		for(int i=0;i<l;i++){
			num[i]=s[i]-'0';
			if(num[i]>3&&num[i]<8) num[i]-=1;//因为没有3
			else if(num[i]>8) num[i]=num[i]-2; 
		}
		int sum=0;
		for(int i=0;i<l;i++)
		 sum=sum*8+num[i];
		 printf("%s: %d\n",s,sum);
		
	}
	return 0;
}