HDOJ-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): 340103    Accepted Submission(s): 65964


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 #define MAXSIZE 1005
 4 
 5 int main()
 6 {
 7     char temp1[MAXSIZE],temp2[MAXSIZE];
 8     int cases;
 9     scanf("%d",&cases);
10     int fuck=cases;
11     while(cases--){
12         int i,j;
13         scanf("%s",temp1);
14         scanf("%s",temp2);
15         int len1=strlen(temp1),len2=strlen(temp2);
16         
17         /*将两个数反转*/ 
18         char val1[MAXSIZE],val2[MAXSIZE];
19         for(i=len1-1,j=0;i>=0;--i)
20             val1[j++]=temp1[i];
21         val1[j]='\0';
22         for(i=len2-1,j=0;i>=0;--i)
23             val2[j++]=temp2[i];
24         val2[j]='\0';
25         
26         /*将长度小的数的高位用0补齐*/ 
27         int max_len=(len1>len2)?len1:len2;
28         if(len1!=len2){
29             
30             if(len1<len2){
31                 for(i=len1;i<len2;++i)
32                     val1[i]='0';
33                 val1[len2]='\0';
34             }
35             else{
36                 for(i=len2;i<len1;++i)
37                     val2[i]='0';
38                 val2[len1]='\0';
39             }
40         }
41         
42         /*每个位做加法（进位）*/ 
43         int flag=0,over=0;
44         char res[MAXSIZE];
45         for(i=0;i<max_len;i++){
46             int sum=val1[i]+val2[i]-2*'0'+flag;
47             if(sum>9){
48                 if(i==(max_len-1))
49                     over=1;
50                 flag=1;    
51                 res[i]=sum-10+'0';
52             }
53             else{
54                 flag=0;
55                 res[i]=sum+'0';
56             }
57         }
58         
59         /*注意考虑第一位进位导致res长度大于max_len的情况*/
60         if(over)
61             res[max_len++]=flag+'0';
62         res[max_len]='\0';
63         
64         /*逐个打印*/ 
65         printf("Case %d:\n",fuck-cases); 
66         printf("%s + %s = ",temp1,temp2);
67         for(i=max_len-1;i>=0;--i)
68             putchar(res[i]);
69         if(cases!=0)
70             printf("\n\n");
71         else
72             putchar('\n');
73     }
74     return 0;
75 }

 

posted @ 2017-01-17 16:08  codinRay  阅读(147)  评论(0)  编辑  收藏  举报