Faulty Odometer(九进制数)

最新推荐文章于 2021-05-17 17:15:16 发布

转载最新推荐文章于 2021-05-17 17:15:16 发布 · 140 阅读

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原文链接：http://www.cnblogs.com/handsomecui/p/6563455.html

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#数据结构与算法

本文介绍了一种特殊情况下里程表的计数问题，该里程表从3直接跳到5，跳过数字4。通过将问题转化为九进制数的处理方式，文章提供了一种有效的解决方案，并附带了实现这一逻辑的C++代码。

Faulty Odometer

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9301		Accepted: 5759

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
题解：油表坏了，遇到4就变成5，其实就是一个九进制数，注意大于4的要减一，我却吓模拟了半天。。。
代码：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>

using namespace std;
int *getary(int N){
    int a[15];
    int *b;
    b = (int *)malloc(sizeof(int) * 15); 
    memset(a, 0, sizeof(a));
    int i = 0;
    while(N){
        a[i ++] = N % 10;
        N /= 10;
    }
    b[0] = i;
    for(int j = 1; j <= i; j++){
        b[j] = a[i - j];
    }
    return b;
}

int main(){
    
    int N;
    while((scanf("%d", &N)), N){
        int *a = getary(N);
        
        int temp = 0;
        for(int i = 1; i <= a[0]; i++){
            if(a[i] > 4){
                a[i] --;
            }
            temp = temp * 9 + a[i];
        }
        printf("%d: %d\n", N, temp);
    }
    
    return 0;
}