Reverse Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5763 Accepted Submission(s): 2643
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
Sample Output
Author
lcy
Source
HDU 2006-4 Programming Contest
感觉用string做比较迅速方便啦…………~~~，这题就该一次提交ac之；
AC code：
#include<iostream>
#include<string>
using namespace std;
int main()
{
int T,n;string str;
while(cin>>T)
{
for(int i=0;i<T;i++)
{
cin>>str;
int len=str.size();int k=len-1;
if(str[0]!='-')
{
for(int i=len-1;i>=0;i--)
if(str[i]!='0')
{
k=i;
break;
}
for(int j=k;j>=0;j--)
cout<<str[j];
for(int p=0;p<len-1-k;p++)
cout<<0;
cout<<endl;
}
else
{
str.erase(0,1);
cout<<'-';
for(int i=len-1-1;i>=0;i--)
if(str[i]!='0')
{
k=i;
break;
}
for(int j=k;j>=0;j--)
cout<<str[j];
for(int p=0;p<len-2-k;p++)
cout<<0;
cout<<endl;
}
}
}
return 0;
}
杭电 HDU 1266 Reverse Number
原文：http://blog.csdn.net/lsgqjh/article/details/44820443
