Sicily 1240. Faulty Odometer-优快云博客

本文探讨了带有特定缺陷的计数器如何工作，并提供了一个示例程序来演示其功能。主要内容包括计数器的工作原理、输入输出示例以及一个用于解决实际问题的程序代码。

1240. Faulty Odometer

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

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

Sample Output

13: 12
15: 13
2003: 1461
2005: 1462
239: 197
250: 198
1399: 1052
1500: 1053
999999: 531440

进制转化题;

#include <stdio.h>
#include <string.h>
#include <math.h>
int main() {
    char num[15];
    while (gets(num)) {
        if (num[0] == '0')
            break;
        printf("%s: ", num);
        int base = pow(9, strlen(num) - 1);
        int sum = 0;
        for (int i = 0; num[i] != '\0'; i++) {
            if (num[i] > '4')
                num[i]--;
            sum += (num[i] - '0') * base;
            base /= 9;
        }
        printf("%d\n", sum);
    }
    return 0;
}

假如是两个数字不能正常显示呢？

#include <stdio.h>
#include <string.h>
#include <math.h>
int main() {
    char num[15];
    while (gets(num)) {
        if (num[0] == '0')
            break;
        printf("%s: ", num);
        int base = pow(8, strlen(num) - 1);
        int sum = 0;
        for (int i = 0; num[i] != '\0'; i++) {
            if (num[i] > '4' && num[i] < '7')
                num[i]--;
            else if (num[i] > '7')
                num[i] -= 2;
            sum += (num[i] - '0') * base;
            base /= 8;
        }
        printf("%d\n", sum);
    }
    return 0;
}