A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 435932    Accepted Submission(s): 84825


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
 

 

 

代码:

 1 #include<stdio.h>
 2 #include<string.h>
 3 int main(){
 4     char a[1005],b[1005];
 5     int c[1005],d[1005],ans[1005];
 6     int n,i,k,j,h,t,l;
 7     int len1,len2;
 8     scanf("%d",&n);
 9         h=0;
10         while(n--){
11             h+=1;
12             for(i=0;i<1005;i++){
13                 c[i]=d[i]=0;
14                 ans[i]=0;
15             }
16             scanf("%s %s",&a,&b);
17             len1=strlen(a);
18             len2=strlen(b);
19             for(i=0;i<len1;i++){
20                c[i]=a[i]-'0';
21             }
22             for(i=0;i<len2;i++){
23                d[i]=b[i]-'0';
24             }
25             printf("Case %d:\n",h);
26             for(i=0;i<len1;i++)
27                 printf("%d",c[i]);
28                 printf(" + ");
29             for(i=0;i<len2;i++)
30                 printf("%d",d[i]);
31                 printf(" = ");
32             for(i=0,j=len1-1;i<len1/2;i++,j--){
33                 t=c[j];
34                 c[j]=c[i];
35                 c[i]=t;
36             }
37             for(i=0,j=len2-1;i<len2/2;i++,j--){
38                 t=d[j];
39                 d[j]=d[i];
40                 d[i]=t;
41             }
42             l=len1>len2?len1:len2;
43             for(i=0,k=0;i<=l;i++,k++){
44                 if(c[i]+d[i]<10)
45                     ans[k]=c[i]+d[i];
46                 else{
47                     ans[k]=(c[i]+d[i])%10;
48                     c[i+1]+=(c[i]+d[i])/10;
49                 }
50 
51             }
52             for(j=k-1;j>=0;j--){
53                 if(j==k-1&&ans[j]==0)
54                     continue;
55                 printf("%d",ans[j]);
56             }
57                 if(n==0)
58                     printf("\n");
59                 else
60                     printf("\n\n");
61         }
62     return 0;
63 }