Reverse Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4022 Accepted Submission(s): 1877
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
Author
lcy
Source
HDU 2006-4 Programming Contest
Recommend
lxj
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<iostream> 5 using namespace std; 6 7 char str[1000]; 8 9 int main() 10 { 11 int n; 12 scanf("%d",&n); 13 getchar(); 14 while(n--) 15 { 16 memset(str,0,sizeof(str)); 17 scanf("%s",str); 18 int len=strlen(str),i; 19 int flag=0; 20 int num=0; 21 int sign=0; 22 for(i=len-1;i>=0;i--) 23 { 24 if(str[0]>'9'||str[0]<'0'&&flag==0) 25 { 26 flag=1; 27 printf("%c",str[0]); 28 } 29 if(flag==1&&i==0)continue; 30 if(str[i]=='0'&&sign==0) 31 { 32 num++; 33 continue; 34 } 35 if(str[i]>='0'&&str[i]<='9') 36 { 37 sign=1; 38 printf("%c",str[i]); 39 } 40 } 41 for(i=0;i<num;i++) 42 printf("0"); 43 printf(" "); 44 } 45 return 0; 46 }