hdu1002 A + B Problem II(大数题)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1002

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

2

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:\n",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 ("\n");

45             if (t)

46                 printf ("\n");

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 }