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
  
  
  
  

   
   ?
  
  
#include<stdio.h>
#include<string.h>
int main()
{
	int t,i,j,k,l,p;
	int q=0;
	char a[1010],b[1010];
	int c[1010]={0},d[1010]={0};
	scanf("%d",&t);
	//getchar();
	while(t--)
	{
		q++;
		memset(a,sizeof(char));
		memset(b,sizeof(char));
		
		<span style="color:#6633ff;BACKGROUND-COLOR: #ffffff">memset(c,sizeof(c));
		memset(d,sizeof(d));
       // memset(c,sizeof(int));
		//memset(d,sizeof(int));       此处错误，sizeof()内是需要清零的数组总长度 
        //memset(c,sizeof(int)*1010);  上下两种均可，切记 ，，切记 ，， 
		//memset(d,sizeof(int)*1010);
</span>		scanf("%s",a);
		scanf("%s",b);
		k=strlen(a);
		l=strlen(b);  
		for(i=k-1,j=0;i>=0;j++,i--)
             c[j]=a[i]-'0';       
		for(i=l-1,i--)
              d[j]=b[i]-'0';    		
		for(i=0;i<1010;i++)
		{
			c[i]=c[i]+d[i];
			if(c[i]>=10)
			{
				c[i]=c[i]%10;c[i+1]++;
			}
		}
		for(i=1010-1;c[i]==0&&i>=0;i--);//第二次做此题，就错在1010没减1，导致正确答案前面有好多0输出。如下部：	
		printf("Case %d:n",q);
		printf("%s + %s = ",a,b);
		if(i>=0)
			for(;i>=0;i--)
			{
				printf("%d",c[i]);
			}
    	else
			printf("0");
		printf("n");
		if(t)
		printf("n");
	}
	return 0;
} 
22222199999990000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000002222210