hdu1002

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 217144    Accepted Submission(s): 41834

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

 

/*大数的相加减问题*/
# include 
# include 
# include 
int main()
{
	int i,j,k,n,z,t,m=1;
	int lena,lenb;
	char a[1001],b[1001];
	int c[1001];
	scanf("%d",&n);
	while(n>=1)
	{
		scanf("%s%s",&a,&b);
		lena=strlen(a)-1;
		lenb=strlen(b)-1;
			k=0;
	        z=0;
		for(;lena>=0&&lenb>=0;lena--,lenb--)
		{
			t=(a[lena]-'0')+(b[lenb]-'0')+z;
			c[k++]=t%10;
			z=t/10;	
		}
		for(;lena>=0;lena--)
		{
			t=(a[lena]-'0')+z;
			c[k++]=t%10;
			z=t/10;
		}
		for(;lenb>=0;lenb--)
		{
			t=(b[lenb]-'0')+z;
			c[k++]=t%10;
			z=t/10;
		}
            printf("Case %d:\n",m++);
            printf("%s + %s = ",a,b);
            for(i=k-1;i>=0;i--)
            {
                printf("%d",c[i]);
            }
            printf("\n");
            if(m<=n)     printf("\n");
            n--;
	}
	return 0;
}