大数A+B

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 314120    Accepted Submission(s): 60873

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

 

Author

Ignatius.L

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
char a[1001],b[1001],p[1001],q[1001];
int max(int x,int y)
{
	int wei=x>y?x:y;
	return wei;
}
int main()
{
	int t,cut=0;
	scanf("%d",&t);

	while(t--)
   { 
       cut++;
     scanf("%s%s",&a,&b);
     	//printf("%s %s
",a,b);
     	memset(q,0,sizeof(q));
     	memset(p,0,sizeof(p));
	int n=strlen(a),m=strlen(b);
	int wei=max(n,m);
		//printf("%d
",wei);
		for(int i=0;i<n;i++)
		{
			q[i]=a[n-1-i]-'0';
		}
			for(int i=0;i<m;i++)
		{
		    p[i]=b[m-1-i]-'0';
		}
		
	for(int i=0;i<wei;i++)
	{
      q[i]=q[i]+p[i];
      if(q[i]>9)
      {
      	q[i]=q[i]%10;
      q[i+1]++;
	  }
      
	}
	printf("Case %d:
",cut);
	printf("%s + %s = ",a,b);
	int j;
	if(q[wei]==0)
     {
	for(j=wei-1;j>=0;j--)
	{
		printf("%d",q[j]);

	}
	printf("
");
    }
	else
	{
			for(j=wei;j>=0;j--)
	{
		printf("%d",q[j]);
	}
	   	    		printf("
");	    
	}
	
	if(t!=0)
	printf("
");
	   
    }
	return 0;
}