对于long long类型也存放不下的超长整数，可以使用字符串模拟的办法来对其进行运算的模拟。此处仅做加法的示范，其他运算思想类似。

每四位视为一个整数，进位等情况对下一次的计算考虑影响。

#include<iostream>
using namespace  std;

#define  MAX_LEN 50
int nNumA[MAX_LEN];
int nLenA = 0;
int nNumB[MAX_LEN];
int nLenB = 0;
int nResult[2*MAX_LEN];
int nLenR = 0;

void CharToInt(char *szSource, int *nDes, int *Len)
{
	int j = 0;
	int nDesLoc = 0;
	int nTotalNum = 0;
	int thisNum = 0; //当前char的值

	int nHead = strlen(szSource) % 4;
	if (nHead != 0)
	{
		for (int k = 0; k< nHead; k++)
		{
			thisNum = (int)(szSource[k] - "0") + 0;
			nTotalNum = nTotalNum*10 + thisNum;
			j++;
	
		}
		nDes[nDesLoc] = nTotalNum;  //数组开头不足四位
		nTotalNum = 0;
		j = 0;
		nDesLoc++;
	}

	for (int i= nHead; i<strlen(szSource); i++)
	{
		if (j == 4)
		{
			nDes[nDesLoc] = nTotalNum;  //每一个int数组为上存储四个数字
			nTotalNum = 0;
			j = 0;
			nDesLoc++;
		}

		thisNum = (int)(szSource[i] - "0") + 0;
		nTotalNum = nTotalNum*10 + thisNum;
		j++;
	}
	if (nTotalNum != 0)
	{
		nDes[nDesLoc] = nTotalNum;   
		nTotalNum = 0;
		j = 0;
		nDesLoc++;
	}
	*Len = nDesLoc;

	cout<<"长度为"<<nDesLoc<<endl;
	for (int j = 0; j< nDesLoc; j++)
	{
		cout<<nDes[j]<<" ";
	}
	cout<<endl;

}

void LenAdd()
{
	int ALen = nLenA;
	int BLen = nLenB;
	int CStart = 0;
	int nFlag = 0;
	while (ALen != 0 && BLen != 0)
	{
		nResult[CStart] = (nNumA[ALen-1] + nNumB[BLen-1] + nFlag) % 9999;
		nFlag = (nNumA[ALen-1] + nNumB[BLen-1] + nFlag) / 9999;
		CStart++;
		ALen--;
		BLen--;

	}

	while (ALen > 0)
	{
		nResult[CStart] = (nNumA[ALen-1]  + nFlag) % 9999;
		nFlag = (nNumA[ALen-1] +  nFlag) / 9999;
		CStart++;
		ALen--;
	}
	while (BLen > 0)
	{
		nResult[CStart] = (nNumB[BLen-1]  + nFlag) % 9999;
		nFlag = (nNumB[BLen-1] +  nFlag) / 9999;
		CStart++;
		BLen--;
	}
	nLenR = CStart;
	cout<<"相加结果为："<<endl;
	for (int i = CStart-1; i >-1 ; i--)
	{
		cout<<nResult[i]<<" ";
	}
	cout<<endl;
}

int main()
{
	char szInput[100];
	cout<<"输入第一个操作数"<<endl;
	cin>>szInput;
	CharToInt(szInput, nNumA, &nLenA);

	cout<<"输入第二个操作数"<<endl;
	cin>>szInput;
	CharToInt(szInput,nNumB, &nLenB);
	LenAdd();
	system("pause");

}