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。

 #include <iostream>
 #include <cstdio>
 #include <cstring>

 using namespace std;

 int main ()
 {
     char a[],b[];
     int na[],nb[],sum[],pre,flag=;
     int t;
     scanf("%d",&t);
         while (t--)
         {
             memset(sum,,sizeof(sum));
             memset(na,,sizeof(na));
             memset(nb,,sizeof(nb));
             scanf("%s%s",a,b);
             pre=;
             int lena=strlen(a);
             int lenb=strlen(b);
             for (int i=; i<lena; i++)
                 na[lena--i]=a[i]-'';
             for (int j=; j<lenb; j++)
                 nb[lenb--j]=b[j]-'';
             int lenx=lena>lenb?lena:lenb;
             for (int k=; k<lenx; k++)
             {
                 sum[k]=na[k]+nb[k]+pre/;
                 pre=sum[k];
             }
             while (pre>)
             {
                 sum[lenx]=pre/%;
                 lenx++;
                 pre/=;
             }
             printf ("Case %d:\n",flag++);
             printf ("%s + %s = ",a,b);
             for (int i=lenx-; i>=; i--)
             {
                 printf ("%d",sum[i]%);
             }
             printf ("\n");
             if (t)
                 printf ("\n");
         }

     return ;
 }

java代码。

 import java.util.*;
 import java.math.*;
 public class Main {
     public static void main(String[] args) {
         Scanner sc=new Scanner (System.in);
         int l=sc.nextInt();
         for(int i=;i<=l;i++){
             if(i!=) System.out.println();
             BigInteger a,b;
             a=sc.nextBigInteger();
             b=sc.nextBigInteger();
             System.out.println("Case "+i+":");
             System.out.println(a+" + "+b+" = "+a.add(b));
         }
     }
 }