杭電1002 A+Bproblem..7回も渡して、最後に5回も削除していないのでテスト出力==、、

A + B Problem II
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 98985    Accepted Submission(s): 18771
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
   
   
   
   

 

  
  
  
  

   
   
   
   
#include<stdio.h> 
   
   
   
   
#include<string.h> 
   
   
   
   
int main() 
   
   
   
   
{ 
   
   
   
   
    int n,i,l1,l2,max,a,j,t; 
   
   
   
   
    char s1[1001],s2[1001],s3[1001]; 
   
   
   
   
    scanf("%d",&n); 
   
   
   
   
    for(i = 0;i < n;i++) 
   
   
   
   
    { 
   
   
   
   
        scanf("%s%s",s1,s2); 
   
   
   
   
 
   
   
   
   
        l1 = strlen(s1); 
   
   
   
   
        l2 = strlen(s2); 
   
   
   
   
        if(l1>l2) 
   
   
   
   
            max = l1; 
   
   
   
   
        else 
   
   
   
   
            max = l2; 
   
   
   
   
 
   
   
   
   
        t=max; 
   
   
   
   
        memset(s3,'\0',1001); 
   
   
   
   
        memset(s3,'0',max+1); 
   
   
   
   
        l1 = l1-1; 
   
   
   
   
        l2 = l2-1; 
   
   
   
   
        max = max; 
   
   
   
   
        s3[0]='0'; 
   
   
   
   
        while(l1>=0 && l2>=0) 
   
   
   
   
        { 
   
   
   
   
            a = s1[l1--]-'0'+s2[l2--]-'0'+s3[max]-'0';//               ，      
   
   
   
   
            s3[max--] ='0'+a%10; 
   
   
   
   
            if(a>=10) 
   
   
   
   
                s3[max]+=1;//            ，               20.。。。 
   
   
   
   
             
   
   
   
   
        } 
   
   
   
   
        while(l1>=0) 
   
   
   
   
        { 
   
   
   
   
            a = s3[max]-'0'+s1[l1--]-'0'; 
   
   
   
   
            s3[max--]='0'+a%10; 
   
   
   
   
            if(a>=10) 
   
   
   
   
                s3[max]+=1; 
   
   
   
   
        } 
   
   
   
   
        while(l2>=0) 
   
   
   
   
        { 
   
   
   
   
            a = s3[max]-'0'+s2[l2--]-'0'; 
   
   
   
   
            s3[max--]='0'+a%10; 
   
   
   
   
            if(a>=10) 
   
   
   
   
                s3[max]+=1; 
   
   
   
   
        } 
   
   
   
   
 
   
   
   
   
        printf("Case %d:
",i+1); 
   
   
   
   
        printf("%s + %s = ",s1,s2); 
   
   
   
   
 
   
   
   
   
        if(s3[0] > '0') 
   
   
   
   
            printf("%c",s3[0]);//      ，   1 9  ，      
   
   
   
   
        for(j = 1;j < t; j++) 
   
   
   
   
            printf( "%c", s3[j]); 
   
   
   
   
        printf( "%c
", s3[j]); 
   
   
   
   
        if(i<n-1) 
   
   
   
   
            printf("
"); 
   
   
   
   
    } 
   
   
   
   
    return 0; 
   
   
   
   
}