A + B Problem II

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

Total Submission(s): 120188    Accepted Submission(s): 22865

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

 

 

1 #include
       
         2 #include
        
          3 #define maxn 1010 4 int shu1[maxn],shu2[maxn]; 5 char str1[maxn],str2[maxn]; 6 int main() 7 { 8     int n,i,j,len1,len2,k; 9     scanf("%d",&n);10     getchar();11     for(k=1;k<=n;k++)12     {13         scanf("%s",str1);14         scanf("%s",str2);15         printf("Case %d:\n%s + %s = ",k,str1,str2);16         memset(shu1,0,sizeof(shu1));17         memset(shu2,0,sizeof(shu2));18         len1=strlen(str1);19         len2=strlen(str2);20         //printf("lenstr1=%d,lenstr2=%d",len1,len2);//21         for(j=0,i=len1-1;i>=0;i--)22             shu1[j++]=str1[i]-'0';23         for(j=0,i=len2-1;i>=0;i--)24             shu2[j++]=str2[i]-'0';25         for(i=0;i
         
          =10)29             {30                 shu1[i]%=10;31                 shu1[i+1]++;32             }33         }34     //for(i=0;i
          
           =0;j--)37           if(shu1[j])break;38        if(j==-1)39            printf("0");40        else 41         {42             for(i=j;i>=0;i--)43            printf("%d",shu1[i]);44         }45      if(k