I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B. 

InputThe 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. 

OutputFor 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

21 2112233445566778899 998877665544332211

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110

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高精度加法模板，注意前导0不输出。

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1 // 高精度加法  2 #include
     
       3 #include
      
        4 using namespace std; 5 int a[1005],b[1005]; 6  7 int pluss(int a[],int b[]){ 8     int l; 9     l=a[0]>b[0]?a[0]:b[0];10     for(int i=1;i<=l;i++){11         a[i+1]+=(a[i]+b[i])/10;12         a[i]=(a[i]+b[i])%10;13     }14     if(a[l+1]>0) a[0]=l+1;15     else a[0]=l;16     return 0;17 }18 19 int main(){20     int T;21     scanf("%d",&T);22     for(int cnt=1;cnt<=T;cnt++){23         24         //读入 25         memset(a,0,sizeof(a));26         memset(b,0,sizeof(b));27         string s1,s2;28         cin>>s1>>s2;29         a[0]=s1.length();30         b[0]=s2.length();31         for(int i=1;i<=a[0];i++){32             a[i]=s1[a[0]-i]-'0';//倒序存储 33         }34         for(int i=1;i<=b[0];i++){35             b[i]=s2[b[0]-i]-'0';36         } 37         38         cout<<"Case "<
       
        <<":"<
        
         <
         
          <<" + "<
          
           <<" = ";39 40 pluss(a,b);41 42 int flag=0;43 for(int i=a[0];i>=1;i--){44 if(flag==0&&a[i]==0) continue;//坑点：前导0不输出。 45 else{46 cout<