Faulty Odometer HDU - 4278 进制问题

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本文介绍了一种特殊里程表的工作原理，该里程表在显示里程时会跳过数字3和8。通过一种转换算法，可以将里程表显示的数字转换为实际行驶的里程数。文章提供了一个C++实现示例，展示了如何逐位处理输入数字并计算出实际里程。

Faulty Odometer
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2087 Accepted Submission(s): 1444

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

15
2005
250
1500
999999
0

Sample Output

15: 12
2005: 1028
250: 160
1500: 768
999999: 262143

Source
2012 ACM/ICPC Asia Regional Tianjin Online

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liuyiding

题意：行车记录仪记录的数会跳过3 跳过8
问行车记录仪实际应该记录的步数
tho: 尝试挺长时间才看出判断每个位置的数然后加起来就行

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
    long long int ave[12];
long long int f(long long int n)
{
    int t;
    int sum=0;
    long long ans=0;
//    if( n%10 > 3 && n%10 <8) ans= n%10- 1;
//    else if(n%10>8) ans = n%10- 2;//最开始写的时候只想起了判个位数，
//    else ans = n%10;//后来改成分开判，最后发现合一块就行。。。
    while(n)
    {
        t=n%10;
        if(t>3&&t<8) t-=1;
        if(t>8) t-=2;
        ans += t*ave[sum];
        sum++;
        n/=10;
    }
    return ans;
}
int main()
{
    int i,j;
    long long int n;
    int t=1;
    ave[0]=0;
    for(i=0;i<11;i++)
    {
        ave[i]=t;
        t*=8;
    }
    while(cin>>n,n!=0)
    {
        long long ans=f(n);
        printf("%d: %d\n",n,ans);
    }
    return 0;
}