POJ Exponentiation解题

Problem: 1001		User: Quincy
Memory: 260K		Time: 0MS
Language: C++		Result: Accepted

• Source Code

// ExponentiationDemo.cpp : Defines the entry point for the console application.

//





#include <string.h>

#include <iostream>



using namespace std;



const int DATA_LENGTH=200;

const int INPUT_LENGTH=100;



struct LargeNumber

{

	int dotIndex;

	int dataLength;

	char data[DATA_LENGTH];

};



struct InputNumber

{

	LargeNumber number;

	int power;

};



void MultiLargeNumber(const LargeNumber* num1, const LargeNumber* num2, LargeNumber* outResult)

{

	int nLength1 = num1->dataLength;

	int nLength2 = num2->dataLength;



	LargeNumber tempValue;

	memset(&tempValue, 0, sizeof(tempValue));



	int ntempValueIndex = 0;



	for (int i=nLength1-1; i >=0; --i, ntempValueIndex++)

	{

		int nCarryNum = 0;

		int index = 0;

		for (int j=nLength2-1; j >= 0; --j, ++index)

		{

			int nData = num1->data[i];

			nData = nData * num2->data[j] + nCarryNum + tempValue.data[index+ntempValueIndex];

			tempValue.data[index+ntempValueIndex] = nData % 10;

			nCarryNum = nData / 10; 

		}



		if (0!=nCarryNum)

		{

			tempValue.data[index+ntempValueIndex] = nCarryNum;

			index++;

		}



		tempValue.dataLength = index + ntempValueIndex;

	}



	tempValue.dotIndex = num1->dotIndex + num2->dotIndex;

	/// Reserve Result to num1;



	memset(outResult, 0, sizeof(LargeNumber));



	int nLengthResult = tempValue.dataLength;   

	for (int i=0; i < nLengthResult; ++i)

	{

		outResult->data[i] = tempValue.data[nLengthResult-i-1];

	}



	outResult->dotIndex = tempValue.dotIndex;

	outResult->dataLength = tempValue.dataLength;

}



bool ConvertStrToNum(const string& str, LargeNumber* num)

{

	int ndataIndex = 0;

	bool bZero = true;

	for(int i=0; i<str.length(); ++i)

	{

		char tempValue = str.at(i);



		if (tempValue == '.')

		{

			num->dotIndex = str.length() - i - 1;

			continue;

		}



		if (tempValue < '0' || tempValue > '9')

		{

			return false;          

		}



		if (tempValue != '0')

		{

			bZero = false;

		}



		num->data[ndataIndex++] = tempValue - '0';

	}



	if (bZero)

	{

		num->dataLength = 0;

		num->dotIndex = 0;

	}

	else

	{

		num->dataLength = ndataIndex;

	}

	

	return true;

}



void PrintLargeNumber(const LargeNumber* num)

{

	char szResult[DATA_LENGTH] = {0};

	int  nResultIndex = 0;



	int nLength = num->dataLength;

	for (int i=0; i<nLength; ++i)

	{

		if (num->dotIndex == nLength-i)

		{

			szResult[nResultIndex++] = '.';

			//strResult.push_back('.');

			//cout << "." ;

		}



		szResult[nResultIndex++] = char(num->data[i]+'0');

		//strResult.push_back(char(num->data[i]+'0'));

		//cout << char(num->data[i]+'0');

	}



	/// Remove back zero



	for (int i = nResultIndex-1; i >= 0; --i)

	{		

		if (szResult[i] != '0')

		{	

			if (szResult[i] == '.')

			{

				szResult[i] = 0;

			}



			break;

		}



		szResult[i] = 0;

	}



	/// Remove front zero



	int nStartPos = 0;

	for (; nStartPos < nResultIndex; ++nStartPos )

	{

		if (szResult[nStartPos] != '0')

		{

			break;

		}

	}



	cout << (char*)(szResult+nStartPos) << endl;

}



int main()

{

	char s[8];

	int n;



	//cin.get(s, 7);

	//cin >> n;

	InputNumber Inputs[INPUT_LENGTH] = {0};

	int nInputCount = 0;

	while(cin >> s >> n)

	{

		if (!ConvertStrToNum(s, &Inputs[nInputCount].number))

		{

			return 0; //continue;

		}



		Inputs[nInputCount].power = n;

		++nInputCount;

	}



	int index=0;

	while (index < nInputCount)

	{

		LargeNumber Result;

		memcpy(&Result, &Inputs[index].number, sizeof(LargeNumber));



		for (int i=0; i<Inputs[index].power-1; ++i)

		{

			MultiLargeNumber(&Inputs[index].number, &Result, &Result);

		}



		PrintLargeNumber(&Result);

		++index;

	}



	return 0;

}

之前在遍历文件夹下的所以文件和文件夹时，我采用的是递归调用的方式。后来发现可以使用非递归的广度优先算法来实现，感觉比自己的代码要好，更优雅些。

因为我很多时候解决问题只是把它解决了就好，没有考虑更换思路,所以决定提高自己的算法方面的知识了，以给自己不同的解决思路。

开始提交几次都出问题了，后来发现还是在现实时把10.0000显示成10.而不是10所以出问题了，主要还是测试用例不全面。