杭电ACM大数加法题目

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基于FPGA的PCIe核系列：FPGA root模式生成工程例子分析（1）一、 PCIE系统架构 1.1 PCIE分层结构 PCIE结构 1.2 TLP包四种TLP类型。 Memory （内存空间操作） IO （IO空间操作） Configuration （配置操作） Message （信息操作）四种类型TLP，如果有需要响应，称为Non-Posted TLP，如果没有需要响应，称为Posted TLP。 TLP包类型 1.2.1 TLP结构 TLP数据格式 1.2.2 Header各个字段的含义 Header格式 ... 阅读详情

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;

	}

}