杭电1266 Reverse Number
Reverse Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3771 Accepted Submission(s): 1736
Problem Description
Welcome to 2006'4 computer college programming contest!
Specially, I give my best regards to all freshmen! You are the future of HDU ACM! And now, I must tell you that ACM problems are always not so easy, but, except this one... Ha-Ha!
Give you an integer; your task is to output its reverse number. Here, reverse number is defined as follows:
1. The reverse number of a positive integer ending without 0 is general reverse, for example, reverse (12) = 21;
2. The reverse number of a negative integer is negative, for example, reverse (-12) = -21;
3. The reverse number of an integer ending with 0 is described as example, reverse (1200) = 2100.
Specially, I give my best regards to all freshmen! You are the future of HDU ACM! And now, I must tell you that ACM problems are always not so easy, but, except this one... Ha-Ha!
Give you an integer; your task is to output its reverse number. Here, reverse number is defined as follows:
1. The reverse number of a positive integer ending without 0 is general reverse, for example, reverse (12) = 21;
2. The reverse number of a negative integer is negative, for example, reverse (-12) = -21;
3. The reverse number of an integer ending with 0 is described as example, reverse (1200) = 2100.
Input
Input
file contains multiple test cases. There is a positive integer n
(n<100) in the first line, which means the number of test cases, and
then n 32-bit integers follow.
Output
For each test case, you should output its reverse number, one case per line.
Sample Input
3
12
-12
1200
Sample Output
21
-21
2100
Author
lcy
Source
Recommend
lxj
如题中描述,这是一道简单题!但开始做的时候考虑的不够周全,遗漏了0出现在整数非末端的情况,导致一次WA!改进后AC,代码如下:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <stdio.h> 2 #include <stdlib.h> 3 4 int reverse( int x ) 5 { 6 int cnt = 0; 7 if( x < 0 ) 8 { 9 printf( "-" ); 10 x = -x; 11 } 12 else if( x == 0 ) 13 { 14 printf( "%d\n",x ); 15 return 0; 16 } 17 while( x % 10 == 0 ) 18 { 19 x = x / 10; 20 cnt++; 21 } 22 while( x != 0 ) 23 { 24 printf( "%d", x % 10 ); 25 x = x / 10; 26 } 27 while( cnt-- ) 28 printf( "%d", 0 ); 29 printf( "\n" ); 30 return 0; 31 } 32 33 int main(int argc, char *argv[]) 34 { 35 int n, a; 36 scanf( "%d", &n ); 37 while( n-- ) 38 { 39 scanf( "%d", &a ); 40 reverse(a); 41 } 42 43 // system("PAUSE"); 44 return 0; 45 }