nyoj103 A+B Problem II （大数加法）

I have a very simple problem for you. Given two integers A and B,your job is to calculate the Sum of A + B.

A,B must be positive.

输入

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.

输出

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.

样例输入

2
1 2
112233445566778899 998877665544332211

样例输出

Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 2222222222222221110

??开始落下了输出格式，赤裸裸的wa。。哈哈

?输入的时候是两个字符串，所以就又开辟了数组，和大数阶乘一样，也是数的个位放在数组num[0]上，注意进位。

代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX 2000
void ADD(char str1[],char str2[])
{
    int len1 = strlen(str1);
    int len2 = strlen(str2);
    //printf("%d %d",len1,len2);
    int num1[MAX];
    int num2[MAX];
    memset(num1,sizeof(num1));
    memset(num2,sizeof(num2));      //注意清0
    int i,j;
    for(i=len1-1,j=0; i>=0; i--)
    {
        num1[j++] = str1[i] - '0';     //注意字符转换成数字的时候要减去‘0’
    }
    for(i=len2-1,j=0; i>=0; i--)
    {
        num2[j++] = str2[i] - '0';
    }
    for(i=0; i<MAX; i++)                
    {
        num1[i] += num2[i];           //开始从个位相加，也就是数组的0开始
        if(num1[i] > 9)               //判断是否产生进位，进位就数组的下一个元素+1
        {
            num1[i] -= 10;
            num1[i+1] ++;
        }
    }
    for(i=MAX-1; i>=0; i--)       //判断第一个不为0的地方，从这里开始输出数组元素，也就是最后的和
    {
        if(num1[i] != 0)
            break;
    }
    printf("%s + %s = ",str1,str2);     //注意输出的结构！！！
    for(j=i; j>=0; j--)
    {
        printf("%d",num1[j]);
    }
    printf("n");
}
int main()
{
    int n;
    scanf("%d",&n);
    getchar();
    int i;
    for(i=1; i<=n; i++)
    {
        char str1[MAX];
        char str2[MAX];
        scanf("%s%s",str2);
        printf("Case %d:n",i);   //注意输出的结构！
        ADD(str1,str2);

    }
    return 0;
}

最优代码

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

    public static void main(String args[]) {
       Scanner cin=new Scanner(System.in);
       int n=cin.nextInt();
       BigInteger a,b;
       for(int i=1;i<=n;i++){
    	   a=cin.nextBigInteger();
    	   b=cin.nextBigInteger();
    	   System.out.println("Case "+i+":");
    	   System.out.println(a.toString()+" + "+b.toString()+" = "+a.add(b));
       }
    }
}
                

额....java 的代码，表示目前还看不懂啊~~