[ACM]A+B Problem(大数加算3つの方法)

Problem Description
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
 
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
 
Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.
 
Sample Input

   
   
   
   

    
    
    
    2
1 2
112233445566778899 998877665544332211
   
   
   
   

 
Sample Output

   
   
   
   

    
    
    
    Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
   
   
   
   

解題の構想:2つの数の各ビットを1つの配列に格納し、逆順序で格納し、対応するビットが加算され、別の配列に格納します.
方法1:

#include <iostream>
#include <string.h>
using namespace std;
int main()
{
    char x[1001],y[1001];//          ，       x  ，       y  
    int xx[1001]={0},yy[1001]={0},sum[1001]={0},i,j=1;//xx  （  ）    x        ，                   ，sum                  
    int lenx,leny,maxlen,t;
    cin>>t;
    while(t--)
    {
        cin>>x>>y;
        lenx=strlen(x);//x   
        leny=strlen(y);//y   
        maxlen=lenx>leny?lenx:leny;//       maxlen
        for(i=0;i<lenx;i++)
            xx[i]=x[lenx-i-1]-'0';//  ascii   '0' 48   '4' 52 ，    4 ，       xx ，  x[lenx-i-1]，        ，          
        for(i=0;i<leny;i++)
            yy[i]=y[leny-i-1]-'0';
        for(i=0;i<maxlen;i++)
        {
            sum[i]+=xx[i]+yy[i];//     
            sum[i+1]=sum[i]/10;//             18，  sum[i]  10， sum[i+1] 1，       1
            sum[i]=sum[i]%10;//         10，  ，    -10；
        }
        if(sum[i]==1)
            maxlen++;//   i  ，          ，       10  ，  1，     s[i]    ，        maxlen+1
        cout<<"Case "<<j<<":"<<endl;
        j++;
        cout<<x<<" + "<<y<<" = ";
        for(i=maxlen-1;i>=0;i--)
            cout<<sum[i];
        cout<<endl;
        if(t!=0)
            cout<<endl;//     ，AC  ，    
        for(i=0;i<maxlen;i++)
        {
            sum[i]=0;
            xx[i]=0;
            yy[i]=0;
        }//  ，         

    }

}

方法2:

#include <iostream>
#include <string.h>
using namespace std;
int main()
{
    char x[1001],y[10001];
    int xx[1001]={0},yy[1001]={0},sum[1001]={0},i,j=1,lenx,leny,maxlen;
    int t;
    cin>>t;
    while(t--)
       {

        cin>>x>>y;
        lenx=strlen(x);
        leny=strlen(y);
        maxlen=lenx>leny?lenx:leny;
        for(i=0;i<lenx;i++)
            xx[i]=x[lenx-i-1]-'0';
        for(i=0;i<leny;i++)
            yy[i]=y[leny-i-1]-'0';
        for(i=0;i<maxlen;i++)
        {
            sum[i]+=xx[i]+yy[i];
            sum[i+1]=sum[i]/10;
            sum[i]=sum[i]%10;//     1  
        }
        if(sum[i]==1)
            maxlen++;
        cout<<"Case "<<j<<":"<<endl;
        j++;
        cout<<x<<" + "<<y<<" = ";
        for(i=maxlen-1;i>=0;i--)
            cout<<sum[i];
        cout<<endl;
        if(t!=0)
            cout<<endl;
        memset(xx,0,1001*sizeof(int));//, xx      0，memset         ，     memset(xx,0,1001);  ，      ，  1001*4， 1001*sizeof(int)
        memset(yy,0,1001*sizeof(int));
        memset(sum,0,1001*sizeof(int));

    }

}

方法3:

#include <iostream>
#include <string.h>
using namespace std;
int main()
{
    char x[1001],y[10001];
    int xx[1001]={0},yy[1001]={0},sum[1001]={0},i,j=1,lenx,leny,maxlen;
    int t;
    cin>>t;
    while(t--)
       {

        cin>>x>>y;
        lenx=strlen(x);
        leny=strlen(y);
        maxlen=lenx>leny?lenx:leny;
        for(i=0;i<lenx;i++)
            xx[i]=x[lenx-i-1]-'0';
        for(i=0;i<leny;i++)
            yy[i]=y[leny-i-1]-'0';
        for(i=0;i<maxlen;i++)
        {
            sum[i]+=xx[i]+yy[i];
            if(sum[i]>=10)//                  10
            {
                sum[i+1]=1;//    10，   1
                sum[i]-=10;//    20，  -10
            }
        }
        if(sum[i]==1)
            maxlen++;
        cout<<"Case "<<j<<":"<<endl;
        j++;
        cout<<x<<" + "<<y<<" = ";
        for(i=maxlen-1;i>=0;i--)
            cout<<sum[i];
        cout<<endl;
        if(t!=0)
            cout<<endl;
        memset(xx,0,1001*sizeof(int));
        memset(yy,0,1001*sizeof(int));
        memset(sum,0,1001*sizeof(int));

    }

}

スクリーンショットの実行: