杭电1002（大数相加）

A + B Problem II

     Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                Total Submission(s): 258229    Accepted Submission(s): 49919

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　　

 

 

//惭愧啊；以前的代码是错的 ；现在修改一下。注意： 字符串以 '' 结尾 ；

#include <stdio.h>
#include <string.h>
char a[1010], b[1010], c[1010] ; 
int main()
{
    
    int i, j, k, n ;
    scanf("%d", &n) ;
    int total = 0 ;
    while(n--)
    {
        memset(c, '', sizeof(c)) ;
        scanf("%s %s", a, b) ;
        int len = strlen(a) ;
        int nel = strlen(b) ;
        if(len > nel)
        {
            for(k = 0, i = len-1, j = nel-1; i >= 0; j--, i--)
            {
                if(j >= 0)
                c[k++] = a[i] + b[j] - '0' ;
                else
                c[k++] = a[i] ;
            }    
        }    
        else
        {
            for(k = 0, i = len - 1, j = nel-1; j >= 0; j--, i--)
            {
                if(i >= 0)
                c[k++] = a[i] + b[j] - '0' ;
                else
                c[k++] = b[j] ;
            }
        }
        for(i = 0; i < k; i++)
        {
            if(c[i] > '9' && i < k-1)
            {
                c[i] -= 10 ;
                c[i+1] += 1 ; 
            }
            if(i == k-1 && c[i] > '9')
            {
                c[i] -= 10 ;
                c[k++] = '1' ;
            } 
        }
            
    //    puts(c) ;
        strrev(c) ;
    //    puts(c) ;
     
        ++total ; 
        printf("Case %d:
", total) ;
        printf("%s + %s = %s
",a, b, c) ; 
        if(n != 0)
        printf("
") ;
        //printf("%d
", n) ;
    }
    return 0 ;
}