hdoj 1002（C实现）

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

#include <stdio.h>
#include <string.h>

int main()
{
	char a;
	int mark,i,j,T,Case = 0,lenA,lenB,lenC;
	char A[1001],B[1001],C[1002];
	if(scanf("%d",&T)!=EOF || T==1 || T==20 || T<20 && T>1)
	{	
		for(j = 1;j < T+1;j ++)
		{
			if(j!=1)          //就是这个点，让我Presentation Error了三次，我的天。太粗心了，以后得注意
				printf("\n");
			mark = 0;
			memset(A,0,1001);
			memset(B,0,1001);
			memset(C,0,1002);
			scanf("%s %s",A,B);
			lenA=strlen(A);
			lenB=strlen(B);
			lenC=0;
			for(i = 0; i < lenA; i++)
				if(A[i]<'0' || A[i]>'9')
				{
					mark = 1;
					break;
				}
			for(i = 0; i < lenB; i++)
				if(B[i]<'0' || B[i]>'9')
				{
					mark = 1;
					break;
				}
			if(mark)
			{
				T++;
				printf("\n");
				continue;
			}
			printf("Case %d:\n%s + %s = ",++Case, A, B);
			for(i = 0; i < lenA/2; i++)
			{
				a = A[i];
				A[i] = A[lenA-i-1];
				A[lenA-i-1] = a;
			}
			for(i = 0; i < lenB/2; i++)
			{
				a = B[i];
				B[i] = B[lenB-i-1];
				B[lenB-i-1] = a;
			}
			i = 0;
			while(i<lenA && i<lenB)
			{
				C[i] += A[i] + B[i] - 96;
				if(C[i] > 9) {
					C[i] = C[i]-10;
					C[i+1] += 1;	
				}
				lenC++;
				i++;
			}
			if(i < lenA)
				while(i <lenA)
				{
					C[i] += A[i] - 48;
					i++;
					lenC++;
				}
			else
				while(i < lenB)
				{
					C[i] += B[i] - 48;
					i++;
					lenC++;
				}
			for(i = lenC-1; i > -1; i--)
				printf("%c",C[i]+48);
			printf("\n");
		}
		
	}
	return 0;
}

VC6运行效果图

*参考资料 http://www.cnblogs.com/JXNU-WuYeqi/p/3485244.html

就是在这里找到了我错误的原因