hduoj 1002 A + B Problem II（大数加法）

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 301097    Accepted Submission(s): 58056

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

AC：java大数

import java.math.BigInteger;
import java.util.Scanner;

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

AC: C大数加法

#include<stdio.h>
#include<string.h>
int main()
{
	char a[1000],b[1000],c[1001];
	int i,j=1,p=0,n,n1,n2;
	scanf("%d",&n);
    while(n)
	{
		scanf("%s %s",a,b);
		printf("Case %d:\n",j);
		printf("%s + %s = ",a,b);
		n1=strlen(a)-1;
		n2=strlen(b)-1;
		for(i=0;n1>=0||n2>=0;i++,n1--,n2--)
		{
			if(n1>=0&&n2>=0){ c[i]=a[n1]+b[n2]-'0'+p;}
			if(n1>=0&&n2<0){ c[i]=a[n1]+p;}
			if(n1<0&&n2>=0){ c[i]=b[n2]+p;}
			p=0;
			if(c[i]>'9'){ c[i]=c[i]-10;p=1;}
		}
		if(p==1){printf("%d",p);}
		while(i--)
		{ printf("%c",c[i]);}
		j++;
		if(n!=1){ printf("\n\n");}
		else {printf("\n");}
		n--;
	}
}