Reverse Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9927 Accepted Submission(s): 4410
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.
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
#include<stdio.h>
#include<string.h>
#define MAXN 10000
char a[MAXN];
int main()
{
int t, n, i, len;
int k;
scanf("%d", &t);
while( t-- )
{
scanf("%s", &a);
len = strlen(a);
for(k = len - 1; k >= 0; k--)
{
if(a[k] != '0')
break;
}
if(a[0] == '-')
{
printf("-");
for(i = k; i >= 1; i--)
printf("%c", a[i]);
for(i = k + 1; i < len; i++)
printf("0");
}
else
{
for(i = k; i >= 0; i--)
printf("%c", a[i]);
for(i = k + 1; i < len; i++)
printf("0");
}
printf("
");
}
return 0;
}