I have a very simple problem for you. Given two integers A and B,your job is to calculate the Sum of A + B.

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.

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.

  
  

   
   2
1 2
112233445566778899 998877665544332211
  
  

  
  

   
   Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 2222222222222221110
  
  
  
  

   
   ?
  
  
  
  

   
   题目大意就是大数相加
  
  
  
  

   
   
#include<stdio.h>
#include<string.h>
int main()
{
	int n,i,j,k,t,lena,lenb,a[1001],b[1001],sum[1001];
	char sa[1001],sb[1001];
	scanf("%d",&n);
	for(k=1;k<=n;++k)
	{
		printf("Case %d:n",k);
		memset(a,sizeof(a));
		memset(b,sizeof(b));
		memset(sum,sizeof(sum));
		scanf("%s %s",sa,sb);
		lena=strlen(sa);
		lenb=strlen(sb);
		for(i=lena-1,j=0;i>=0;j++,i--)a[j]=sa[i]-'0';
		for(i=lenb-1,i--)b[j]=sb[i]-'0';
		for(i=0;i<1001;++i)
		{
			sum[i]=sum[i]+a[i]+b[i];
			if(sum[i]>=10)
			{
				sum[i]-=10;
				sum[i+1]+=1;
			}
		}
		printf("%s + %s = ",sb);
		i=1000;
		while(sum[i]==0&&i>=0)i--;
		for(;i>=0;i--)printf("%d",sum[i]);
		printf("n");
		if(k!=n)printf("n");
	}
	return 0;
}