hdoj 1002 A + B Problem II【大数加法】

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 260585    Accepted Submission(s): 50389

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>
#define MAX 1100
char str1[MAX],str2[MAX];
int a[MAX],b[MAX];
int main()
{
	int n,m,j,i,s,t,l1,l2,k,ok;
	scanf("%d",&t);
	k=1;
	while(t--)
	{
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		scanf("%s%s",str1,str2);
		l1=strlen(str1);
		l2=strlen(str2);
		for(i=l1-1,j=0;i>=0;i--)
		{
			a[j]=str1[i]-'0';
			j++;
		}
		for(i=l2-1,j=0;i>=0;i--)
		{
			b[j]=str2[i]-'0';
			j++;
		}
		n=0;
		if(l1<l2)
		{
			n=l1;
			l1=l2;
			l2=n;
		}
	    for(i=0;i<l1;i++)
	    {
	    	a[i]=a[i]+b[i];
	    	if(a[i]>=10)
	    	{
	    		a[i]-=10;
	    		a[i+1]++;
	    	}
	    }
	    ok=0;
	    if(a[l1]>0)
	    {
	    	ok=1;
	    }
	    printf("Case %d:\n",k++);
	    printf("%s + %s = ",str1,str2);
	    if(!ok)
	    for(i=l1-1;i>=0;i--)
	    printf("%d",a[i]);
	    else
	    for(i=l1;i>=0;i--)
	    printf("%d",a[i]);
	    printf("\n");
	    if(t)
	    printf("\n");
	}
	return 0;
}