HDU 4278 Faulty Odometer【进制转换】

Faulty Odometer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2017    Accepted Submission(s): 1398

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

 

Recommend

liuyiding   |   We have carefully selected several similar problems for you:  4267 4268 4269 4270 4271 

 

比较明显的8进制转10进制的题目，只要处理一下因为跳过数字产生的误差就可以了。

#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <string>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
//#define LOACL
#define space " "
using namespace std;
typedef long long LL;
//typedef __int64 Int;
typedef pair<int, int> PAI;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double PI = acos(-1.0);
const int MOD = 1e9 + 7;
const int MAXN = 50 + 10;
int bit[MAXN];
int main() {
    int N;
    while (scanf("%d", &N), N) {
        int cnt = 0;
        int t = N;
        while (N) {
            bit[cnt++] = N%10;
            N /= 10;
        }
        for (int i = 0; i < cnt; i++) {
            if (bit[i] > 3) bit[i]--;
            if (bit[i] == 8) bit[i]--;
        }
        int ans = 0;
        for (int i = 0; i < cnt; i++) {
            ans += bit[i]*pow(8, i);
        }
        printf("%d: %d
", t, ans);
    }
    return 0;
}