HDU--1002 -- A + B Problem II

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A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 165473 Accepted Submission(s): 31614

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

Code:

#include"stdio.h"
#include"string.h"
int main()
{
	char num1[1001],num2[1002];
	int n1[1002],n2[1002];
	int T,ca,i,j,l1,l2,l3;
	scanf("%d",&T);
	for(ca=1;ca<=T;ca++)
	{
		
		memset(n1,0,sizeof(n1));//全部置为0 
		memset(n2,0,sizeof(n2));
		scanf("%s%s",num1,num2);
		l1 = strlen(num1);
		l2 = strlen(num2);
		l3 = l1>l2?l1:l2;
				
		printf("Case %d:\n",ca);//输出 
		for(i=0;i<l1;i++)
			printf("%c",num1[i]);
		printf(" + ");
		for(i=0;i<l2;i++)
			printf("%c",num2[i]);
		printf(" = ");
		
		for(i=0,j=l1-1;i<l1,j>=0;j--)//字符串逆置 
			n1[i++] = num1[j] - '0';
		for(i=0,j=l2-1;i<l2,j>=0;j--)
			n2[i++] = num2[j] - '0';	
		
		for(i=0;i<1001;i++)	//相加 
		{
			n1[i] = n1[i] + n2[i];
			if(n1[i]>9)
			{
				n1[i] = n1[i] - 10;
				n1[i+1]++;
			}
		}		
		if(n1[l3]) l3++;//若最高位有进位,则输出个数加1 <l3为最大长度,有进位说明长度加1>
		for(i=l3-1;(i>=0 && n1[i]=='0');i--);
		if(i>=0)
			for(;i>=0;i--)
				printf("%d",n1[i]);
		else printf("0");//0+0时空循环中的i变为-1,需另考虑 

		printf("\n");
		if(ca!=T)	printf("\n");
	}
}

要比较长度啊!! 啊啊啊啊