[C++日常小题] Find Actual Number of Miles-优快云博客

本文链接：https://blog.youkuaiyun.com/Elliott_Yoho/article/details/53458250

这篇博客讨论了一个具有缺陷的汽车里程表问题，该里程表在从2跳到4和从7跳到9时会跳过数字3和8。作者给出了输入和输出的详细描述，包括如何处理从含有缺陷的里程读数转换为实际行驶的英里数的方法。示例输入和输出展示了如何将里程表读数15和250转换为实际的行驶距离12和160英里。

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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

250

Sample Output

15: 12

250: 160

Code

#include <cstdio>
#include <cmath>
int main() {
    int n, m, i, tem, count, num, temp;
    while (scanf("%d", &n) == 1 && n != 0) {
        count = 0, m = n, num = 1;
        for (i = 8; i >= 0; --i)
            if (n >= pow(10, i)) break;
        for (int j = 0; j < i; ++j) num *= 10;
        while (i >= 0) {
            tem = num = 1;
            for (int j = 0; j < i; ++j) {
                tem *= 8;
                num *= 10;
            }
            temp = n / num;
            if (temp > 7) {
                temp -= 2;
            } else if (temp > 2) {
                temp--;
            }
            count += temp * tem;
            n %= num;
            i--;
        }
        printf("%d: %d\n", m, count);
    }
    return 0;
}