大数相加

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 <stdio.h>

 2 #include <string.h>

 3 int main()

 4 {

 5 

 6         int x, sa[1001], sb[1001], k = 1, i, j, m, n, t, d;

 7         char a[1001], b[1001];

 8         scanf("%d", &x);

 9         while( x-- ){

10 

11                 //int sa[1001] = {0};

12                 //int sb[1001] = {0};#1

13                 memset(sa, 0, sizeof(sa));

14                 memset(sb, 0, sizeof(sb));

15                 scanf("%s", a);

16                 scanf("%s", b);

17                 m = strlen(a);

18                 n = strlen(b);

19                 for( i = 0, j = m - 1; j >= 0; j-- )

20                 {

21 

22                         //sa[i++] = a[j];

23                         sa[i++] = a[j] - '0';

24 

25                 }

26                 for( i = 0, j = n - 1; j >= 0; j-- )

27                 {

28 

29                         //sb[i++] = b[j];

30                         sb[i++] = b[j] - '0';

31                 }

32                 t = m >= n ? m : n;/*找出最长的长度*/

33                 for( i = 0; i < t; i++ )

34                 {

35 

36                         sa[i] += sb[i];

37                         if( sa[i] >= 10 )

38                         {

39 

40                                 sa[i] %= 10;    

41                                 sa[i + 1]++;

42 

43                         }

44 

45                 }

46                 //d = (sa[i] == 1) ? t : t - abs(m - n) ;

47                 d = (sa[i] == 1) ? t : t - 1;/*如果最高位还进1位，那就从它的上位开始输出*/

48                 printf("Case %d:\n%s + %s = ", k++, a, b);

49                 while( d >= 0 ){

50 

51                         //printf("%d", d--);

52                           printf("%d", sa[d--]);    

53 

54                 }

55                 putchar('\n');

56                 //putchar('\n');细节问题：最后一次测试无空行

57                 //return 0;

58                 if( x == 0 )

59                         continue;

60                 putchar('\n');

61 

62         }

63         return 0;

64 

65 }

View Code

 

 

note:

#1只有定义变量才可以全部赋值，但只能是0。一般不喜欢这样做，所以调用头文件<string.h>里的memset函数，它不仅可以全盘赋值，而且赋值自由。

#2 细节问题：因为Output a blank line between two test cases，所以最后一组数据后面没有空行