杭电1266--数的倒叙

                                        Reverse Number

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

代码1：

# include <iostream>
# include <cstdio>
# include <algorithm>
# include <cstring>

using namespace std;

void f(char  *p,int i,int j);

int main(){

    char st[100];
    int i,j,len;
    int n,m;

    while(scanf("%d",&n)!=EOF){

        for(i=0;i<n;i++){
            memset(st,0,sizeof(st));
            scanf("%s",st);

            if(st[0]!='-'){
                len = strlen(st);

                j= len - 1;
                while(st[j]=='0'&&j!=0) j--;

             f(st,0,j);
            }else{
                len = strlen(st);
                j= len - 1;
                while(st[j]=='0'&&j!=0) j--;

             f(st,1,j);
            }


            printf("%s\n",st);




        }

    }


    return 0;
}



void f(char *p,int i,int j){

    char t;
    int n=(i+j)/2;
    while(i<=n){
        t = p[i];
        p[i] = p[j];
        p[j] = t;
        i++;
        j--;

    }
}

代码2：

#include <iostream>
using namespace std;
int main()
{
    int n;
    cin>>n;
    while(n--)
    {
        _int64 x,t,flag;
        int count=0;
        cin>>x;
        t=0;
        flag=0;
        if(x<0)
        {
            x=-1*x;
            flag=1;
        }
        while(x>0)
        {

            t=t*10+x%10;
            x/=10;
            if(!t)
                count++;
        }    

        if(flag)
            cout<<'-';
        cout<<t;    
        while(count--)
        {
            cout<<0;
        }
        cout<<endl;
    }
    return 0;
}