HDUoj 4278 Faulty Odometer （数学

最新推荐文章于 2019-09-24 21:19:39 发布

Yishui_Blog

最新推荐文章于 2019-09-24 21:19:39 发布

阅读量287

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/wang2332/article/details/61952159

online judge HDU 专栏收录该内容

110 篇文章

订阅专栏

本文介绍了一种特殊的里程表，该里程表从2跳到4，从7跳到9，忽略数字3和8。通过输入里程表显示的数值，转换为实际行驶的里程数，并提供了解决方案的代码实现。

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): 2017 Accepted Submission(s): 1398

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

就是跳过3与8，想了很长时间参考了下别人的代码

#include <stdio.h>
#include <string.h>
#define LL long long 
char str[100];
int main()
{
    while(~scanf("%s",str)) {
        if(str[0] == '0') {
            break;
        }
        int len = strlen(str);
        LL ans = 0, k = 1;
        for(int i = len-1;i >= 0; i--) {
            LL x = str[i] - '0';
            if(x == 9) {
                x = x-2;
            }
            else if(x >= 4) 
                x = x-1;
            ans += x*k;
            k *= 8;
        }  
        printf("%s: %lld\n",str,ans);
    }
return 0;
}