Reverse Number
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Reverse Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3872 Accepted Submission(s): 1784
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
312-121200
Sample Output
21-212100
#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>using namespace std ;int main(void){ int T; cin >> T ; while(T--){ int n; cin >> n ; char str[102]; sprintf(str,"%d",n); int flag = 0 ; if(n<0) cout<<"-" ; reverse(str,str+strlen(str)); int cnt = 0 , i , j ; for( i = 0 ; i < strlen(str) ; i++) if(str[i]=='0') cnt ++ ; else break; for( ; i < strlen(str) ; i ++) if(str[i]>='0'&&str[i]<='9') cout<<str[i]; for( j = 0 ; j < cnt ; j ++) cout<<"0"; cout<<endl; } return 0;}
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