HDOJ1002-A + B Problem II(高精加)

Posted 2020-09-19

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

Means:

求大整数的A+B

Solve:

高精模板题

Code:

 1 //仅当两个都为正
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstring>
 5 using namespace std;
 6 
 7 struct big_number
 8 {
 9     int digit[10000];
10     int len;
11     big_number()
12     {
13         memset(digit,0,sizeof(digit));
14         len=0;
15     }
16 };
17 
18 big_number reset_number(char *in)//将低位存在数组前面，如123存到数组里就是321
19 {
20     big_number after;
21     after.len=strlen(in);
22     for(int i=0;i<after.len;++i)
23     {
24         after.digit[i]=(in[after.len-i-1]-48);
25     }
26     return after;
27 }
28 
29 big_number add(big_number num1,big_number num2)
30 {
31     int carry=0;
32     big_number ans;
33     for(int i=0;i<num1.len || i<num2.len;++i)
34     {
35         int temp=num1.digit[i]+num2.digit[i]+carry;
36         ans.digit[ans.len++]=temp%10;
37         carry=temp/10;
38     }
39     if(carry!=0)
40         ans.digit[ans.len++]=carry;
41     return ans;
42 }
43 
44 int main()
45 {
46 
47     int t;
48     scanf("%d" , &t);
49     for(int c = 1 ; c <= t ; ++c)
50     {
51         //printf("%s\n%s\n",in1,in2);
52         char in1[10000]={‘\0‘},in2[10000]={‘\0‘};
53         scanf("%s%s",in1,in2);
54         big_number change1=reset_number(in1);
55         big_number change2=reset_number(in2);
56         big_number ans=add(change1,change2);
57         printf("Case %d:\n" , c);
58         printf("%s + %s = " , in1 , in2);
59         for(int i=ans.len-1;i>=0;--i)
60         {
61             printf("%d",ans.digit[i]);
62         }
63         if(c != t)
64             printf("\n\n");
65         else
66             printf("\n");
67     }
68     return 0;
69 }

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