HDOJ1002题A + B Problem II，2个大数相加

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

#include < stdio.h>
#include < stdlib.h>
#include < string.h>
int main()
{ char a[1010],b[1010],c[1010]; int a1,b1,m,i,l,i1,j1,n1,m1,a2,b2,j=1,t1,t2,n,p=0; scanf("%d",&n); while(n--) { p=0; scanf("%s",a); scanf("%s",b); printf("Case %d:\n",j++); printf("%s + %s = ",a,b); a1=strlen(a); b1=strlen(b); a2=a1; b2=b1; for(i=0; a1>=0||b1>=0; i++,a1--,b1--) { if(a1>=0&&b1>=0) { c[i]=a[a1]+b[b1]-'0'+p; } else if(a1>=0&&b1<0) { c[i]=a[a1]+p; } else if(a1<0&&b1>=0) { c[i]=b[b1]+p ; } p=0; if(c[i]>'9') { c[i]=c[i]-10; p=1; } } if(p==1) printf("%d",p); t1=1; t2=i-1; n1=m1=0; for(i1=0; i1 { if(a[i1]=='0') n1++; } for(j1=0 ; j1 { if(b[j1]=='0') m1++; } if(n1==a2&&m1==b2) { printf("0"); } else { for(l= i-1 ; l>0; l--) { if(t2==l&&c[l]=='0'&&p!=1) { t2--; continue; } printf("%c",c[l]); } } if(n!=0) printf("\n\n"); else printf("\n"); } return 0;
}
  

 

  
1
  
2
  
3
  
4
  
5
  
6
  
7
  
8
  
9
  
10
  
11
  
12
  
13
  
14
  
15
  
16
  
17
  
18
  
19
  
20
  
21
  
22
  
23
  
24
  
25
  
26
  
27
  
28
  
29
  
30
  
31
  
32
  
33
  
34
  
35
  
36
  
37
  
38
  
39
  
40
  
41
  
42
  
43
  
44
  
45
  
46
  
47
  
48
  
49
  
50
  
51
  
52
  
53
  
54
  
55
  
56
  
57
  
58
  
59
  
60
  
61
  
62
  
63
  
64
  
65
  
66
  
67
  
68
  
69
  
70
  
71
  
72
  
73
  
74
  
75
  
76
  
77
  
78
 

文章来源: chenhx.blog.csdn.net，作者：谙忆，版权归原作者所有，如需转载，请联系作者。

原文链接：chenhx.blog.csdn.net/article/details/48751909