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
Recommend
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; }