hdoj 1002（C实现）

我会带着你远行 2022-08-14 05:43 150阅读 0赞

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

![20151124132342308][]

\*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

[20151124132342308]: /images/20220731/aceb5d319e4544a89e10eb30a8a86e40.png