A - A + B Problem II

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概述

A - A + B Problem II

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

思路：对于这种大数问题，按位取数，字符串s1和s2，代表输入的数据，把字符串分别倒着存储到a和b，加法运算倒着相加，max是用来标记长度最长的字符串，但是考虑到进位的问题，所以max要多一步判断。

加法准则，从后往前加。

注：提交多次格式错误，特别注意cnt的数据，它的初始数据为1，cnt代表第几组，当cnt！=T的时候，就代表要多输出一行空白，所以要多一步判断，更加要注意的是cnt++的位置，很容易出错。

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<stdlib.h>
#include<string.h>
using namespace std;
char s1[10000],s2[10000];
int a[10000],b[10000],c[10000];
int main(){
	int T,i,j;
	cin>>T;
	int cnt=1;
	for(int k=0;k<T;k++){
		scanf("%s %s",s1,s2);
		memset(c,0,sizeof(c));
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		int len1=strlen(s1);
		int len2=strlen(s2);
		for(i=0;i<len1;i++){
			a[i]=s1[len1-i-1]-'0';  //把数字一个一个的倒着存储在数组中
		}
		for(i=0;i<len2;i++){
			b[i]=s2[len2-i-1]-'0';
		}
		int max=len1>len2?len1:len2;
		for(i=0;i<=max;i++){
			int sum=a[i]+b[i]+c[i];
			if(sum>=10){//如果两个数字相加大于10，就证明要进位
				c[i]=sum-10;//c数组存放进位后的数字
				c[i+1]++;//进位后，c数组下一位的数据要加1
			}else{
				c[i]=sum;//如果不用进位的话，直接把相加后的结果存在c数组中，方便输出
			}
		}
		printf("Case %d:n",cnt); 
		printf("%s + %s = ",s1,s2);
		if(c[max]==0){
			max--;  //前导0的判断
		}
		for(i=max;i>=0;i--){
			cout<<c[i];
		}
		if(cnt!=T)
		cout<<endl<<endl; //考虑到每两组数据间多一行空格
		else
		cout<<endl; 
		cnt++;
	}
	return 0;
}

Java代码

注：类名必须为Main，不能为其他，不然容易出错；包也不用导入，导入也出错

import java.util.Scanner;
import java.math.BigInteger;
public class Main {
   public static void main(String args[]){
	   int cnt=0;
	Scanner scan=new Scanner(System.in);
	   int n=scan.nextInt();
	   while(n!=0){
		   n--;
		   BigInteger a,b;
		   a=scan.nextBigInteger();
		   b=scan.nextBigInteger();
		   System.out.println("Case "+(++cnt)+":");
		   System.out.println(a+" + "+b+" = "+a.add(b));
		   if(n!=0)
			   System.out.println();
	   }
   }
}