hdu1002 A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 413509    Accepted Submission(s): 80134

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

#include<stdio.h>
#include<string.h>
int main()
{ 
	int N;
	int count=1;
	scanf("%d",&N);
	while(N--)
	{
		char a[1000],b[1000],c[1001];
		int i,p=0,n1,n2;
		scanf("%s %s",a,b);
		printf("Case %d:\n",count++);
		printf("%s + %s = ",a,b);
		n1=strlen(a)-1;//如输入12345,串长是5,但是最后一位5在数组里的下标是4 
		n2=strlen(b)-1;
		for(i=0;n1>=0||n2>=0;i++,n1--,n2--)//注意出来的结果是倒序的 
		{   
			if(n1>=0&&n2>=0){c[i]=a[n1]+b[n2]-'0'+p;}//当两者位数相同
					if(n1>=0&&n2<0){c[i]=a[n1]+p;}//位数不同时会出现 
			if(n1<0&&n2>=0){c[i]=b[n2]+p;}
			p=0;
			if(c[i]>'9'){c[i]-=10;p=1;}//进位
		}	
		if(p==1) 
		printf("%d",p);//如果进位，先输入1
		while(i--)//巧妙利用上面循环中的变量i 细微处见真功夫 
		printf("%c",c[i]);
		if(N!=1) 
		printf("\n\n");//Output a blank line between two test cases.
		else printf("\n");//特别注意 只有在两个输出间才需要多输一行，最后一个输出完时只需一个\n 
	}
 	return 0;	
}