今天上午看了一下榜，http://acmicpc.info/icpc2012/t…，发现大家 a 题刷刷的过，于是自己也敲了一下，好久没有敲题，居然一次过了 sample，呵呵，不过后来跟同学讨论的时候，发现我那个暴力模拟的思路实在太弱智了，在同学的指点下改成了进制转换的思路，没有账号，所以没有的交，不过 sample 是过了

题目，附上 pdf，2012TianJinOnlineProblemSet

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

代码

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main() {
	int x;
	while (scanf("%d", &x) && x!=0) {
		char str[15] = {0};
		itoa(x, str, 10);
		for (int i=0; i<strlen(str); i++) {
			if (str[i] >= '8') {
				str[i] -= 2;
				continue;
			}
			if (str[i] >= '3') {
				str[i] -= 1;
				continue;
			}
		}
		int ans = 0;
		for (char *p=str; *p != '\0'; p++) {
			ans = ans*8 + (*p-'0');
		}
		printf("%d: %dn", x, ans);
	}
	return 0;
}

输入输出

15
15: 12
2005
2005: 1028
250
250: 160
1500
1500: 768
999999
999999: 262143
0

===================================

2012-9-9 21:30:09 update oj 上开放了题目，http://acm.hdu.edu.cn/showprob…，于是交了一下，不能用 itoa，于是改成了这样，能够 ac

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main() {
	int x;
	while (scanf("%d", &x) && x!=0) {
		char str[15] = {0};
		sprintf(str, "%d", x);
		for (int i=0; i<strlen(str); i++) {
			if (str[i] >= '8') {
				str[i] -= 2;
				continue;
			}
			if (str[i] >= '3') {
				str[i] -= 1;
				continue;
			}
		}
		int ans = 0;
		for (char *p=str; *p != '\0'; p++) {
			ans = ans*8 + (*p-'0');
		}
		printf("%d: %dn", x, ans);
	}
	return 0;
}