HDUOJ1002A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 120188    Accepted Submission(s): 22865

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

 

 

View Code

 1 #include
 2 #include<string.h>
 3 #define maxn 1010
 4 int shu1[maxn],shu2[maxn];
 5 char str1[maxn],str2[maxn];
 6 int main()
 7 {
 8     int n,i,j,len1,len2,k;
 9     scanf("%d",&n);
10     getchar();
11     for(k=1;k<=n;k++)
12     {
13         scanf("%s",str1);
14         scanf("%s",str2);
15         printf("Case %d:\n%s + %s = ",k,str1,str2);
16         memset(shu1,0,sizeof(shu1));
17         memset(shu2,0,sizeof(shu2));
18         len1=strlen(str1);
19         len2=strlen(str2);
20         //printf("lenstr1=%d,lenstr2=%d",len1,len2);//
21         for(j=0,i=len1-1;i>=0;i--)
22             shu1[j++]=str1[i]-'0';
23         for(j=0,i=len2-1;i>=0;i--)
24             shu2[j++]=str2[i]-'0';
25         for(i=0;i)
26         {
27             shu1[i]+=shu2[i];
28             if(shu1[i]>=10)
29             {
30                 shu1[i]%=10;
31                 shu1[i+1]++;
32             }
33         }
34     //for(i=0;i35         //printf("%d",shu1[i]);//
36        for(j=maxn-1;j>=0;j--)
37           if(shu1[j])break;
38        if(j==-1)
39            printf("0");
40        else 
41         {
42             for(i=j;i>=0;i--)
43            printf("%d",shu1[i]);
44         }
45      if(k<n)
46      printf("\n\n");
47      else
48      printf("\n");
49     }
50 return 0;
51 }

 

转载于:https://www.cnblogs.com/zhaojiedi/archive/2012/07/27/2613432.html