hdu 1002 A + B Problem II 解题报告

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1002

 　 　　　　　　　　　　　　　　A + B Problem II 　

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
  
    

　这也是一个大数加法的模板

  
    

   
      1 
   
     #include
   
     <
   
     stdio.h
   
     >
   
     

   
      2 
   
     #include
   
     <
   
     string
   
     .h
   
     >
   
     

   
      3 
   
      
   
     #define
   
      Max 10000
   
     

   
      4 
   
     
   
     char
   
      sun[Max],str1[Max],str2[Max];

   
      5 
   
     
   
     void
   
      jia( )

   
      6 
   
     {

   
      7 
   
      memset( sun,
   
     0
   
     ,
   
     sizeof
   
     ( sun ) );

   
      8 
   
      
   
     int
   
      len1 
   
     =
   
      strlen( str1 ),len2 
   
     =
   
      strlen( str2 );

   
      9 
   
      
   
     for
   
     ( 
   
     int
   
      i 
   
     =
   
      
   
     0
   
     ; i 
   
     <
   
      ( len1 
   
     >
   
      len2 
   
     ?
   
      len1 : len2 ); 
   
     ++
   
     i )
   
     //
   
     转换成ASCII码
   
     

   
     10 
   
     
   
      str1[i] 
   
     -=
   
      
   
     '
   
     0
   
     '
   
     ,str2[i] 
   
     -=
   
      
   
     '
   
     0
   
     '
   
     ;

   
     11 
   
      
   
     for
   
     ( 
   
     int
   
      p 
   
     =
   
      
   
     0
   
     , q 
   
     =
   
      len1 
   
     -
   
      
   
     1
   
     ; p 
   
     <
   
      q; 
   
     ++
   
     p , 
   
     --
   
     q ) 
   
     //
   
     逆序
   
     

   
     12 
   
     
   
      {

   
     13 
   
      
   
     char
   
      c 
   
     =
   
      str1[q];

   
     14 
   
      str1[q] 
   
     =
   
      str1[p];

   
     15 
   
      str1[p] 
   
     =
   
      c;

   
     16 
   
      }

   
     17 
   
      
   
     for
   
     ( 
   
     int
   
      p 
   
     =
   
      
   
     0
   
     , q 
   
     =
   
      len2 
   
     -
   
      
   
     1
   
     ; p 
   
     <
   
      q; 
   
     ++
   
     p,
   
     --
   
     q)

   
     18 
   
      {

   
     19 
   
      
   
     char
   
      c 
   
     =
   
      str2[q];

   
     20 
   
      str2[q] 
   
     =
   
      str2[p];

   
     21 
   
      str2[p] 
   
     =
   
      c;

   
     22 
   
      }

   
     23 
   
      
   
     for
   
     ( 
   
     int
   
      i 
   
     =
   
      
   
     0
   
     ,c 
   
     =
   
      
   
     0
   
     ; i 
   
     <
   
      ( len1 
   
     >
   
      len2 
   
     ?
   
      len1 : len2 ) 
   
     ||
   
      c; 
   
     ++
   
     i )
   
     //
   
     相加 进位
   
     

   
     24 
   
     
   
      {

   
     25 
   
      
   
     if
   
     ( i 
   
     <
   
      len1 )

   
     26 
   
      c 
   
     +=
   
      str1[i];

   
     27 
   
      
   
     if
   
     ( i 
   
     <
   
      len2 )

   
     28 
   
      c 
   
     +=
   
      str2[i];

   
     29 
   
      sun[i] 
   
     =
   
      c 
   
     %
   
      
   
     10
   
     ;
   
     //
   
     不同的地方1
   
     

   
     30 
   
     
   
      c 
   
     /=
   
      
   
     10
   
     ;
   
     //
   
     不同的地方2
   
     

   
     31 
   
     
   
      }

   
     32 
   
      }

   
     33 
   
     
   
     int
   
      main(){

   
     34 
   
      
   
     int
   
      T;

   
     35 
   
      scanf( 
   
     "
   
     %d
   
     "
   
     ,
   
     &
   
     T );

   
     36 
   
      
   
     int
   
      m
   
     =
   
     T;

   
     37 
   
      
   
     while
   
     ( T
   
     --
   
      ){

   
     38 
   
      scanf( 
   
     "
   
     %s%s
   
     "
   
     ,str1,str2 );

   
     39 
   
      
   
     char
   
      s1[Max],s2[Max];

   
     40 
   
      strcpy( s1,str1 );

   
     41 
   
      strcpy( s2,str2 );

   
     42 
   
      jia( );

   
     43 
   
      printf( 
   
     "
   
     Case %d:\n
   
     "
   
     ,m
   
     -
   
     T );

   
     44 
   
      printf( 
   
     "
   
     %s + %s = 
   
     "
   
     ,s1,s2 );

   
     45 
   
      
   
     int
   
      n
   
     =
   
     Max;

   
     46 
   
      
   
     while
   
     ( 
   
     !
   
     sun[
   
     --
   
     n] );

   
     47 
   
      
   
     for
   
     ( ; n 
   
     >=
   
      
   
     0
   
     ; 
   
     --
   
     n )

   
     48 
   
      printf( 
   
     "
   
     %d
   
     "
   
     ,sun[n]);
   
     //
   
     不同的地方3
   
     

   
     49 
   
     
   
      puts( 
   
     ""
   
      );

   
     50 
   
      
   
     if
   
     (T
   
     >
   
     0
   
     )

   
     51 
   
      puts( 
   
     ""
   
      );

   
     52 
   
      }

   
     53 
   
     }