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  • Faulty Odometer(九进制数)

    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

    13
    15
    2003
    2005
    239
    250
    1399
    1500
    999999
    0
    

    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, temp);
        }
        
        return 0;
    } 


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  • 原文地址:https://www.cnblogs.com/handsomecui/p/6563455.html
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