A+B长字符串_you may assume the length of each integer will not-优快云博客

本文介绍了一种处理大数加法的算法实现方法。针对传统32位整数处理大数时的局限性，该算法使用字符数组存储大数，并通过逐位相加的方式计算两个大数的和。文章提供了完整的C语言代码示例，适用于解决输入包含多个测试用例的大数加法问题。

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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




#include <stdio.h>
#include <string.h>
#define N 1001
char A[N],B[N],sum[N];
int main()
{
	int T,i,j,k,x,carry;
    scanf("%d",&T);
	
		for(i=0;i<T;i++)
		{
			if(i)
			printf("\n");
			scanf("%s%s",&A,&B);
			j=strlen(A)-1;
			k=strlen(B)-1;
			for(x=0,carry=0;(j+1)&&(k+1)!=0;j--,k--,x++)
			{
				if((A[j]-'0')+(B[k]-'0')+carry<10)
				{
					sum[x]=(A[j]-'0')+(B[k]-'0')+carry;
					carry=0;
				}
				else
				{
					sum[x]=(A[j]-'0')+(B[k]-'0')+carry-10;
					carry=1;
				}
			}
			if(j+1)
			{
				for(;j>=0;j--,x++)
				{
					if(A[j]-'0'+carry<10)
					{
						sum[x]=(A[j]-'0')+carry;
						carry=0;
					}
					else
					{
						sum[x]=0;
						carry=1;
					}
				}
			}
			else if(k+1)
			{
				for(;k>=0;k--,x++)
				{
					if(B[k]-'0'+carry<10)
					{
						sum[x]=(B[k]-'0')+carry;
						carry=0;
					}
					else{
						sum[x]=0;
						carry=1;
					}
				}
			 } 
			 if(carry)
			 sum[x]=1;
			 else
			 x--;
			 printf("Case %d:\n",i+1);
			 printf("%s + %s = ",A,B);
			 while(x>-1)
			 printf("%d",sum[x--]);
			 printf("\n");
		}
	
	return 0;
 }