A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 305856    Accepted Submission(s): 59088

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

 

 

Author

Ignatius.L

 

 

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#include
       
        #include
        
         #include
         
          using namespace std;int main(){    int n;    string a,b;    while(~scanf("%d",&n))    {        int cas=1;        for(int i=0; i
          
           >a>>b;            int ans[10005];            for(int i=0; i<1005; i++)                ans[i]=0;            int tmp=0;            int len=(a.length()
           
            b.length())?a.length():b.length();            int t=0;            /* for(int i=llen-1, j=len-1; j>-1 ; i--,j--)             {                 if(a[i]-'0'+b[j]-'0'+tmp>9)                 {                     ans[t++]=a[i]-'0'+b[j]-'0'+tmp-10;                     tmp=1;                 }                 else                 {                     ans[t++]=a[i]-'0'+b[j]-'0'+tmp;                     tmp=0;                 }             }*/            if(a.length()>=b.length())            {                for(int i=llen-1, j=len-1; j>-1 ; i--,j--)                {                    if(a[i]-'0'+b[j]-'0'+tmp>9)                    {                        ans[t++]=a[i]-'0'+b[j]-'0'+tmp-10;                        tmp=1;                    }                    else                    {                        ans[t++]=a[i]-'0'+b[j]-'0'+tmp;                        tmp=0;                    }                }                for(int i=llen-len-1; i>-1; i--)                {                    if(tmp+a[i]-'0'>9)                    {                        ans[t++]=tmp-10+a[i]-'0';                        tmp=1;                    }                    else                    {                        ans[t++]=a[i]+tmp-'0';                        tmp=0;                    }                }            }            if(a.length()
            
             -1 ; i--,j--) { if(a[i]-'0'+b[j]-'0'+tmp>9) { ans[t++]=a[i]-'0'+b[j]-'0'+tmp-10; tmp=1; } else { ans[t++]=a[i]-'0'+b[j]-'0'+tmp; tmp=0; } } for(int i=llen-len-1; i>-1; i--) { if(tmp+b[i]-'0'>9) { ans[t++]=tmp-10+b[i]-'0'; tmp=1; } else { ans[t++]=b[i]+tmp-'0'; tmp=0; } } } if(tmp!=0) ans[t++]=tmp; cout<<"Case "<
             
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