A + B Problem II（杭电1002）

/*A + B Problem II
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*/

/*借鉴大神经验，与新手共勉。

 

*/

#include<stdio.h>
#include<string.h>
int main()
{
    int str1[1100],str2[1100];
    char a[1100],b[1100];
    int test,i,j,t,k=1,len1,len2;
    scanf("%d",&test);
    getchar();
    while(test--)
    {
        char sum[1100]={0};
        scanf("%s %s",a,b);
        getchar();
        len1=strlen(a);
        len2=strlen(b);
        memset(str1,0,sizeof(str1));
        memset(str2,0,sizeof(str2));
        for(i=0,j=len1-1;i<len1;i++,j--)
        str1[j]=a[i]-'0';
        for(i=0,j=len2-1;i<len2;i++,j--)
        str2[j]=b[i]-'0';
        if(len1<len2)
        {
            t=len2;
            len2=len1;
            len1=t;
        }
        for(i=0;i<=len1;i++)
        {
            sum[i]=str1[i]+str2[i]+sum[i];
            if(sum[i]/10>0)
            {
                sum[i+1]=sum[i]/10;
                sum[i]=sum[i]%10;
            }
        }
        printf("Case %d:\n",k);
        k++;
        printf("%s + %s = ",a,b);
        if(sum[len1]!=0)
        printf("%d",sum[len1]);
        for(j=len1-1;j>=0;j--)
        printf("%d",sum[j]);
        printf("\n");
        if(test!=0)
        printf("\n");
    }
    return 0;
}

今天尝试用java写这道大数题目，贴下代码：

import java.math.BigInteger;
 import java.util.Scanner;
 public class Main {


public static void main(String[] args) {

Scanner in=new Scanner(System.in);
int test=in.nextInt();
String s1,s2;
int kase=1;
while(test-->0)
{
s1=in.next();
s2=in.next();
BigInteger s3=new BigInteger(s1);
BigInteger s4=new BigInteger(s2);
System.out.println("Case "+kase+++":");
System.out.println(s3+" + "+s4+" = "+s3.add(s4));
if(test!=0)
System.out.println();
}
}


 }