nyoj103 A+B Problem II （大數加法）

A+B Problem II

時間限制：3000 ms  |  內存限制：65535 KB

難度：3

描述

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.

輸入

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.

輸出

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.

樣例輸入

2
1 2
112233445566778899 998877665544332211

樣例輸出

Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

  開始落下了輸出格式，赤裸裸的wa。。哈哈

 輸入的時候是兩個字符串，所以就又開闢了數組，和大數階乘一樣，也是數的個位放在數組num[0]上，注意進位。

代碼：

#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, 0, sizeof(num1));
    memset(num2, 0, sizeof(num2));      //注意清0
    int i, j;
    for(i=len1-1, j=0; i>=0; i--)
    {
        num1[j++] = str1[i] - '0';     //注意字符轉換成數字的時候要減去‘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];           //開始從個位相加，也就是數組的0開始
        if(num1[i] > 9)               //判斷是否產生進位，進位就數組的下一個元素+1
        {
            num1[i] -= 10;
            num1[i+1] ++;
        }
    }
    for(i=MAX-1; i>=0; i--)       //判斷第一個不爲0的地方，從這裏開始輸出數組元素，也就是最後的和
    {
        if(num1[i] != 0)
            break;
    }
    printf("%s + %s = ", str1, str2);     //注意輸出的結構！！！
    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", str1, str2);
        printf("Case %d:\n", i);   //注意輸出的結構！
        ADD(str1, str2);

    }
    return 0;
}

最優代碼

 
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));
       }
    }
}
                

額....java 的代碼，表示目前還看不懂啊~~