hdu4278 Faulty Odometer 八进制转十进制

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本文介绍了一个带有特殊缺陷的汽车里程表，该缺陷导致它跳过数字3和8。文章提供了一个算法，用于从里程表读数中推算实际行驶的里程。输入为里程表读数，输出为实际行驶里程。

Faulty Odometer

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1875 Accepted Submission(s): 1295

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

Sample Output

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

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

int ans[50], cnt;

void cc(int n) {
	cnt = 0;
	do {
		ans[cnt++] = n % 10;
		n /= 10;
	} while (n);
	for (int i = 0; i < cnt; i++) {
		if (ans[i] > 3 && ans[i] < 8) {
			ans[i] -= 1;
		}
		else if (ans[i] == 9) {
			ans[i] -= 2;
		}
	}
}

int main()
{
	int n;
	while (~scanf("%d", &n) && n) {
		cc(n);
		int sum = 0;
		for (int i = cnt - 1; i >= 0; i--) {
			sum *= 8;
			sum += ans[i];
		}
		printf("%d: %d\n", n, sum);
	}
	return 0;
}