hdoj1002(大数相加)

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

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 119434 Accepted Submission(s): 22702

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

套用模版做的。。。

#include<stdio.h>
#include<string.h>
char str1 [ 10000 ],str2 [ 10000 ];
char sum [ 100001 ];
void jiafa ()
{
      memset (sum , 0 , sizeof (sum ));
    int i ,j ,p ;
    int len1 =strlen (str1 ),len2 =strlen (str2 );
    for (i = 0 ;i <(len1 >len2 ?len1 :len2 );i ++)
    {
              str1 [i ]-= '0' ;
              str2 [i ]-= '0' ;
    }
    for (i = 0 ,j =len1 - 1 ;i <j ;i ++,j --)
    {
        char c =str1 [j ];
              str1 [j ]=str1 [i ];
              str1 [i ]=c ;
    }
    for (i = 0 ,j =len2 - 1 ;i <j ;i ++,j --)
    {
        char c =str2 [j ];
              str2 [j ]=str2 [i ];
              str2 [i ]=c ;
    }
    char c ;
    for (i = 0 ,c = 0 ;i <(len1 >len2 ?len1 :len2 )||c ;i ++)
    {
        if (i <len1 )
                      c +=str1 [i ];
        if (i <len2 )
                      c +=str2 [i ];
              sum [i ]=c % 10 ;
              c /= 10 ;
    }
    for (i =(len1 >len2 ?len1 :len2 )- 1 ;i >= 0 ;i --)
              printf ( "%d" ,sum [i ]);
      printf ( "\n" );
}
int main ()
{
    int n ;
      scanf ( "%d" ,&n );
    int count = 1 ;
    while (n --)
    {
              scanf ( "%s %s" ,str1 ,str2 );
              printf ( "Case %d:\n" ,count ++);
              printf ( "%s + %s = " ,str2 );
              jiafa ();
        if (n )
                      printf ( "\n" );
    }
    return 0 ; }