喜极而泣！杭电OJ——1002 A + B Problem II

A + B Problem II

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

在杭电上交了12次，终于AC，掩饰不住心中的激动！特发此文，以表纪念！

这是一道大数相加的题目，测试数据已经远远超过了基本数据的表示范围，要用数组一位位相加！

以下是AC代码：

仅供参考

//设想：建立一个数组储存加的值
//经验总结：细心，细心，再细心！不然太容易出错！！！
#include<iostream>
#include<string>
#include<cstdio>
using namespace std;

int main()
{
	string a,b;
	int num,sum[1000];//数组尽量开大点，不然很容易越界！
    cin>>num;
	for(int l=0;l<num;l++)
	{
		cin>>a>>b;
		int m=0,n=0,temp=0,i,k;
		int	q=0;//用于记录sum数组的长度！！
	    i=a.length();
	    k=b.length();
		//for(i=0;a[i]!='\0';i++);
	//for(k=0;b[k]!='\0';k++);
		i=i-1;
		k=k-1;
       // cout<<"刚开始时 i="<<i<<"  "<<"k="<<k<<endl;
		while(i>=0 && k>=0)//像是这样加，会一直加到一个数加完
		{
			m=a[i]-'0';
		//	cout<<"m="<<m<<" ";
			n=b[k]-'0';
		//	cout<<"n="<<n<<endl;
		
			sum[q++]=(temp+m+n)%10;
		//	cout<<"sum[q-1]="<<sum[q-1]<<endl;
			temp=(temp+m+n)/10;//temp取进位，并且加到下一次的加法中
			//cout<<"temp="<<temp<<endl;
		//	cout<<"q="<<q<<endl;
			i--;
			k--;
		//	cout<<"i="<<i<<" "<<"k="<<k<<endl;
		//	cout<<endl;
		}
		//还需要处理一个长度的问题，判断哪个长，哪个短！
	  if(i>k)//a数比较大，确切的说是比较长！
	  {//cout<<"I am 1"<<endl;
		  while(i>=0)
		  {
		     
			 m=a[i]-'0';
			 sum[q++]=(temp+m)%10;
		     temp=(temp+m)/10;//坑爹的坑，在这里坑了一次，我本来是temp+m+n，然后不知为何错了，很难察觉的！

			 i--;
		  }
	  }
	  if(k>i)
	  {
		 // cout<<"I am 2"<<endl;
		// cout<<"k="<<k<<" "<<"temp="<<temp<<endl;
		 while(k>=0)
		  {
		     m=b[k]-'0';//这里也被坑了一次，我把b写成了a，全是复制粘贴惹的祸
			//cout<<"m="<<m<<endl;
			 sum[q++]=(temp+m)%10;
            //cout<<"sum[q-1]="<<sum[q-1]<<endl;
		     temp=(temp+m)/10;
			 k--;
		  }
	  }
	  sum[q]=temp;
	 // for(i=50;;i--)
		 // if(sum[i]!=0)
		//	  break;
		  //for(i++;i>=0;i--)
	  cout<<"Case "<<l+1<<":"<<endl;
	 cout<<a<<" + "<<b<<" = ";
	  if(sum[q]!=0) cout<<sum[q];
		  for(--q;q>=0;q--)
          cout<<sum[q];
	 //printf("%d",sum[--q]);//第一个测试数据成功！
	// for(--q;q>=0;q--)
	// {
	//	 printf("%04d",sum[q]);
	// }
	 cout<<endl;
	 if(l<num-1) cout<<endl;

	}//for
 return 0;
}