大数相加 C++版+java版

深碍√TFBOYSˉ_ 2024-02-19 19:24 9阅读 0赞

## A + B Problem II ##

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 468200 Accepted Submission(s): 90552**

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

### 要注意格式啊，每两组测试用例之间有一空行，也就是说最后一组例子没空行。 ###

自己写的：

很容易理解，模拟两个数相加

#include<iostream>
    #include<cstring>
    #include<cstdio>
    using namespace std;
    #define range 1005
    int m=1;
    void add(char a[],char b[],char c[])
    {
    	int i,j,k=0,current,carry=0,len;
    	len=strlen(a)>strlen(b)?strlen(a):strlen(b);
    	len+=3;
    	char res[range];
    	char x,y;
    	i=strlen(a)-1;
    	j=strlen(b)-1;
    	while(i>=0||j>=0)
    	{
    		if(i<0)
    		x='0';
    		else
    		x=a[i];
    		if(j<0)
    		y='0';
    		else
    		y=b[j];
    		current=x-'0'+y-'0';
    		if(carry)
    		current+=1;
    		if(current>9)
    		{
    			current%=10;
    			carry=1;
    		}
    		else carry=0;   //这忘了一次，哇了两次 
    		res[k++]=current+'0';
    		i--;
    		j--;
    	}
    	if(carry)
    	res[k++]='1';
    	res[k]='\0';
    	
    	for(i=k-1,j=0;i>=0;i--,j++)
    	c[j]=res[i];
    	c[j]='\0';
    	return ;
    }
    int main()
    {
    	char a[range],b[range],c[range];
    	int n;
    	cin>>n;
    	while(n--)
    	{
    		cin>>a>>b;
    		add(a,b,c);
    		
    		cout<<"Case "<<m<<":"<<endl;
    		cout<<a<<" + "<<b<<" = "<<c<<endl;
    		if(n!=0)
    		cout<<endl;
    		m++;
    	}
    	return 0;
    }

java版的：

开挂吧

package 大数;
    
    import java.math.BigDecimal;
    import java.util.Scanner;
    
    public class Main {
    
    	public static void main(String[] args) {
    		// TODO Auto-generated method stub
    		Scanner in=new Scanner(System.in);
    		int n=in.nextInt();
    		int m=1;
    		while(n>0)
    		{
    			BigDecimal a=in.nextBigDecimal();
    			BigDecimal b=in.nextBigDecimal();
    			System.out.println("Case "+m+":");
    			System.out.println(a+" + "+b+" = "+a.add(b));
    			if(n!=1)
    				System.out.println();
    			
    			n--;
    			m++;
    		}
    
    	}
    
    }

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