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[1005],b1[1005];
    int a[1005],b[1005],c[1005];//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=1;j<=n;j++)
        {
            memset(a,0,sizeof(a));//数组a清零
            memset(b,0,sizeof(b));//数组b清零
            memset(c,0,sizeof(c));//数组c清零
            scanf("%s%s",&a1,&b1);
            lena=strlen(a1);
            lenb=strlen(b1);
            for(i=0;i<=lena;i++)
                a[lena-i]=a1[i]-'0';//将数串a1转化为数组a，并倒序存储
            for(i=0;i<=lenb;i++)
                b[lenb-i]=b1[i]-'0';//将数串b1转化为数组a，并倒序存储
            x=0;//x是进位
            lenc=1;//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]/10;//向高位进位
                c[lenc]%=10;//存储第lenc位的值
                lenc++;//位置下标变量
            }
            c[lenc]=x;
            if(c[lenc]==0)//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>=1;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 0;
}

 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));
               }
           }
       }
    }
}