1002 A + B Problem II

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 o
utput 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

#include<iostream>
#include <algorithm>
#include <cstring>
#include <stdio.h>
using namespace std;

int main()
{
    int n,c[1005],k,l,sum;
    char a[1005],b[1005],e[2005],f[2005];
    while(cin>>n)
    {
        for(int j=1;j<=n;j++)
        {
            memset(e,'0',sizeof(e));
            memset(f,'0',sizeof(f));
            memset(c,0,sizeof(c));
            cin>>a>>b;
            k=strlen(a);l=strlen(b);
            strcpy(e+1000,a);
            strcpy(f+1000,b);
            for(int i=0;i<max(k,l);i++)
            {
                sum=(e[k-i+999]-'0')+(f[l-i+999]-'0')+c[i];
                c[i]=sum%10;
                c[i+1]=sum/10;
            }
            printf("Case %d:
",j);
            printf("%s + %s = ",a,b);
            if(c[max(k,l)]!=0)
            cout<<c[max(k,l)];
            for(int i=max(k,l)-1;i>=0;i--)
            cout<<c[i];
            cout<<endl;
            if(j!=n)
            cout<<endl;
        }

    }
    return 0;
}

当前面的数对后面有影响时可以，转换变量来消除这种影响。