A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 230247 Accepted Submission(s): 44185

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

1 2

112233445566778899 998877665544332211

Sample Output

Case 1:

1 + 2 = 3

Case 2:

112233445566778899 + 998877665544332211 = 1111111111111111110

题目大意：题意很容易理解，具体就不解释了，主要就是要解决大数的问题。

题目思路：如果会java的话，可以轻松AC。其他的小伙伴们只能用最笨的方法解决。我们用一个数字将数字倒过来存下，无论是乘法还是加法，这是最好的解决办法。

下面附上两个代码。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 
 5 using namespace std;
 6 
 7 int main ()
 8 {
 9     char a[8000],b[8000];
10     int na[8000],nb[8000],sum[8000],pre,flag=1;
11     int t;
12     scanf("%d",&t);
13         while (t--)
14         {
15             memset(sum,0,sizeof(sum));
16             memset(na,0,sizeof(na));
17             memset(nb,0,sizeof(nb));
18             scanf("%s%s",a,b);
19             pre=0;
20             int lena=strlen(a);
21             int lenb=strlen(b);
22             for (int i=0; i<lena; i++)
23                 na[lena-1-i]=a[i]-'0';
24             for (int j=0; j<lenb; j++)
25                 nb[lenb-1-j]=b[j]-'0';
26             int lenx=lena>lenb?lena:lenb;
27             for (int k=0; k<lenx; k++)
28             {
29                 sum[k]=na[k]+nb[k]+pre/10;
30                 pre=sum[k];
31             }
32             while (pre>9)
33             {
34                 sum[lenx]=pre/10%10;
35                 lenx++;
36                 pre/=10;
37             }
38             printf ("Case %d:
",flag++);
39             printf ("%s + %s = ",a,b);
40             for (int i=lenx-1; i>=0; i--)
41             {
42                 printf ("%d",sum[i]%10);
43             }
44             printf ("
");
45             if (t)
46                 printf ("
");
47         }
48 
49     return 0;
50 }

java代码。

 1 import java.util.*;
 2 import java.math.*;
 3 public class Main {
 4     public static void main(String[] args) {
 5         Scanner sc=new Scanner (System.in);
 6         int l=sc.nextInt();
 7         for(int i=1;i<=l;i++){
 8             if(i!=1) System.out.println();
 9             BigInteger a,b;
10             a=sc.nextBigInteger();
11             b=sc.nextBigInteger();
12             System.out.println("Case "+i+":");
13             System.out.println(a+" + "+b+" = "+a.add(b));
14         }
15     }
16 }