HDU 1002 A + B Problem II(高精度加法(C++/Java))

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 347161    Accepted Submission(s): 67385

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

Author

Ignatius.L

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

分析：高精度计算，大数相加！模版在博客中已给出，翻翻看，按照模版写就行了，要注意细节，空格的输出，因为这个PE了2次！

下面给出AC代码：

 #include <bits/stdc++.h>
 using namespace std;
 int main()
 {
     char a1[],b1[];
     int a[],b[],c[];//a，b，c分别存储加数，加数，结果
     int x,i,j,n;
     int lena,lenb,lenc;
     while(scanf("%d",&n)!=EOF)
     {
         for(j=;j<=n;j++)
         {
             memset(a,,sizeof(a));//数组a清零
             memset(b,,sizeof(b));//数组b清零
             memset(c,,sizeof(c));//数组c清零
             scanf("%s%s",&a1,&b1);
             lena=strlen(a1);
             lenb=strlen(b1);
             for(i=;i<=lena;i++)
                 a[lena-i]=a1[i]-'';//将数串a1转化为数组a，并倒序存储
             for(i=;i<=lenb;i++)
                 b[lenb-i]=b1[i]-'';//将数串b1转化为数组a，并倒序存储
             x=;//x是进位
             lenc=;//lenc表示第几位
             while(lenc<=lena||lenc<=lenb)
             {
                 c[lenc]=a[lenc]+b[lenc]+x;//第lenc位相加并加上次的进位
                 x=c[lenc]/;//向高位进位
                 c[lenc]%=;//存储第lenc位的值
                 lenc++;//位置下标变量
             }
             c[lenc]=x;
             if(c[lenc]==)//处理最高进位
                 lenc--;
             printf("Case %d:\n",j);//格式要求吧！学着点
              printf("%s + %s = ",a1,b1);//这也是格式要求吧！学着点
             for(i=lenc;i>=;i--)
             printf("%d",c[i]);
             printf("\n");
             if(j!=n)//对于2组之间加空行的情况
             printf("\n");
         }
     }
     return ;
 }

java写法大数，真是有毒！

 import java.math.BigInteger;
 import java.util.Scanner;

 public class Main {

     /**
      * @param args
      */
     public static void main(String[] args)
     {
         // TODO Auto-generated method stub
        //System.out.println("Hello World!");
        Scanner in=new Scanner(System.in);
        while(in.hasNextInt())
        {
 //           int []arr=new int[3];
            int  n;
            n=in.nextInt();
            for(int i=1;i<=n;i++)
            {
                BigInteger a,b;
                a=in.nextBigInteger();
                b=in.nextBigInteger();
                if(i<n)
                {
                    System.out.println("Case "+i+":");
                    System.out.print(a+" + "+b+" = ");
                    System.out.println(a.add(b));
                    System.out.println();
                }
                else
                {
                    System.out.println("Case "+i+":");
                    System.out.print(a+" + "+b+" = ");
                    System.out.println(a.add(b));
                }
            }
        }
     }
 }