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[8000],b[8000];
    int na[8000],nb[8000],sum[8000],pre,flag=1;
    int t;
    scanf("%d",&t);
        while (t--)
        {
            memset(sum,0,sizeof(sum));
            memset(na,0,sizeof(na));
            memset(nb,0,sizeof(nb));
            scanf("%s%s",a,b);
            pre=0;
            int lena=strlen(a);
            int lenb=strlen(b);
            for (int i=0; i<lena; i++)
                na[lena-1-i]=a[i]-'0';
            for (int j=0; j<lenb; j++)
                nb[lenb-1-j]=b[j]-'0';
            int lenx=lena>lenb?lena:lenb;
            for (int k=0; k<lenx; k++)
            {
                sum[k]=na[k]+nb[k]+pre/10;
                pre=sum[k];
            }
            while (pre>9)
            {
                sum[lenx]=pre/10%10;
                lenx++;
                pre/=10;
            }
            printf ("Case %d:
",flag++);
            printf ("%s + %s = ",a,b);
            for (int i=lenx-1; i>=0; i--)
            {
                printf ("%d",sum[i]%10);
            }
            printf ("
");
            if (t)
                printf ("
");
        }

    return 0;
}

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=1;i<=l;i++){
            if(i!=1) 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));
        }
    }
}