A + B Problem II-优快云博客

本文介绍了一种处理非常大数相加问题的算法，通过数组按位计算实现了两个大数的相加，并展示了如何输出结果。

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
分析：此题就是求两数相加求和，但此题是特殊的大数相加，所以很容易想到数组，按位计算；
代码：#include<stdio.h>
#include<string.h>
int main()
{
	int a[1010],b[1010],sum[1015]; 
	char a1[1010],b1[1010];
	int p1,p2,k=0,i,j,n,max,t;
	scanf("%d",&n);
	while(n--)
	{
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		memset(sum,0,sizeof(sum));
		scanf("%s%s",a1,b1);
     	p1=strlen(a1);
		p2=strlen(b1);
		for(i=0,j=p1-1;j>=0;i++,j--)
		a[j]=a1[i]-'0';
		for(i=0,j=p2-1;j>=0;j--,i++)
		b[j]=b1[i]-'0';
	//	max=(p1>p2)?p1:p2;
	  if(p1<p2)
	  {
       t=p2;
	    p2=p1;	  	
	  	p1=t;
	  }
		for(i=0;i<p1;i++)
	{
		sum[i]+=a[i]+b[i];
		if(sum[i]>9)
		{
			sum[i+1]++;
			sum[i]-=10;
		}
	}
		printf("Case %d:\n",++k);
		printf("%s + %s = ",a1,b1);
		if(sum[p1]!=0)
		printf("%d",sum[p1]);
		for(i=p1-1;i>=0;i--)
		printf("%d",sum[i]);
		printf("\n");
		if(n>0)
		printf("\n");
			}
	return 0;
}