杭电ACM大数加法题目

[ruanjianchjian]

A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 124156 Accepted Submission(s): 23887


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
我写的代码：请各位帮忙看一下为什么是Wrong Answer……

#include<iostream>
#include<string>
using namespace std;
#define Max 1000
int num1[Max];
int num2[Max];

struct bigadd
{
	string str1;
	string str2;
	int * result;
	int len;
};

int main()
{
	string str1,str2;
	int i;
	bool flag=false;
	int casenumber,len1,len2,len,hold=0;
	bigadd * arr;
	memset(num1,0,sizeof(num1));
	memset(num2,0,sizeof(num2));
	cin>>casenumber;
	arr = new bigadd[casenumber];
	for(i=0;i<casenumber;i++)
	{
		cin>>str1>>str2;
		arr[i].str1=str1;
		arr[i].str2=str2;
		len1=str1.length();
		len2=str2.length();
		for(int k=len1-1;k>=0;k--)
		{
			num1[len1-1-k]=str1[k]-'0';
		}
		for(int j=len2-1;j>=0;j--)
		{
			num2[len2-1-j]=str2[j]-'0';
		}
		len=len1>len2?len1:len2;
		arr[i].len=len;
		arr[i].result=new int[len+1];
		for(int k=0;k<=len;k++)
		{
			num1[k]+=num2[k]+hold;
			if(num1[k]>=10)
			{
				hold=num1[k]/10;
				num1[k]=num1[k]%10;
			}
			else
				hold=0;
		}
		for(int l=0;l<=len;l++)
		{
			arr[i].result[l]=num1[l];
		}
		memset(num1,0,sizeof(num1));
		memset(num2,0,sizeof(num2));
	}
	for( i=0;i<casenumber;i++)
	{
		cout<<"Case"<<" "<<i+1<<":"<<endl;
		cout<<arr[i].str1<<" "<<"+"<<" "<<arr[i].str2<<" "<<"="<<" ";
		for(int j=arr[i].len;j>=0;j--)
		{
			if(arr[i].result[j]||flag)
			{
				flag=true;
				cout<<arr[i].result[j];
			}
			
		}
		cout<<endl<<endl;
	}
}

[FancyMouse]

这组数据：
2
11 11
1 1
你flag没有对于每个数据初始化false。

另外
Output a blank line between two test cases.
可能最后一条后面不应该输出endl。

[LogicTeamLeader]

难道不能用java中的BigInteger么？

[Quote=引用楼主 的回复:]
A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 124156 Accepted Submission(s): 23887


Problem Description
I have a very si……
[/Quote]

[skillart]

以前用c的字符串模拟，后来又用java的大数累，真的吊爆了。

[litiouslove]

[Quote=引用 1 楼 的回复:]

这组数据：
2
11 11
1 1
你flag没有对于每个数据初始化false。

另外
Output a blank line between two test cases.
可能最后一条后面不应该输出endl。
[/Quote]
最后一行是要输出endl的，中间是要输出两个endl的。
需要输出一个空行嘛，自然得两个endl，另外杭电的题貌似最后都得endl。
楼主也在搞ACM吗？我刚刚入门，小菜一个，一起努力呗？

[litiouslove]

3
123 1
4 2
11 9
Case 1:
123 + 1 = 124

Case 2:
4 + 2 = 06

Case 3:
11 + 9 = 020

这组数据...首位多了0

[ruralboy1]

用字符数组处理，这题貌似是杭电1002还是1003的吧