A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 177301    Accepted Submission(s): 33930

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

//大数相加
#include<iostream>
#include<string>
#define MAX 1002
using namespace std;

int main()
{
    int T,c[MAX],i,j,k,alen,blen,clen,cnt,carry,sum;
    string a,b;
    cin>>T;
    cnt=0;
    while(cnt<T)
    {
        cnt++;
        cin>>a>>b;
        alen = a.length();
        blen = b.length();
        if(alen>blen)clen=blen;
        else clen = alen;
        for(i=alen-1,j=blen-1,k=0,carry = 0;k<clen;k++,i--,j--)
        {
            sum = a[i]+b[j]+carry-96;
            if(sum > 9){carry=1;sum-=10;}
            else carry = 0;
            c[k]=sum;
        }
        for(;i>-1;i--,k++)
        {
            sum = a[i]+carry-48;
            if(sum > 9){carry = 1;sum -= 10;}
            else carry = 0;
            c[k] = sum;
        }
        for(;j>-1;j--,k++)
        {
            sum = b[j] + carry - 48;
            if(sum > 9)
            {
                carry = 1;
                sum -= 10;
            }
            else carry = 0;
            c[k] = sum;
        }
        if(carry){c[k]=1;k++;}

        cout<<"Case "<<cnt<<":"<<endl;
        cout<<a<<" + "<<b<<" = ";
        for(i=k-1;i>-1;i--)
            cout<<c[i];
        if(cnt!=T)cout<<"

";
        else cout<<"
";
    }
    system("pause");
    return 0;
}