POJ 1001-Exponentiation,测试用例及本人未优化代码

最新推荐文章于 2022-04-30 23:01:46 发布

tianyuan521521

最新推荐文章于 2022-04-30 23:01:46 发布

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分类专栏： C++语言特性文章标签： ACM POJ c++ 测试

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该博客主要介绍了POJ 1001题目的解决方案，需要计算0.0 < R < 99.999之间实数R的n次方，其中0 < n <= 25。博主给出了未优化的C++代码，内存占用512KB，运行时间为16MS。

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Problems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of the national debt is a taxing experience for many computer systems.

This problem requires that you write a program to compute the exact value of Rⁿ where R is a real number ( 0.0 < R < 99.999 ) and n is an integer such that 0 < n <= 25.

测试用例：

注意结果的前后零都不能有，但允许出现 .123这样的但是不能有 1000.这样的，就是说小数点可以是第一位，但不能是最后一位。

代码为c++代码，比较渣，内存占用512K,时间为16MS,未优化：

#include<iostream>
#include<string>
using namespace std;

string BigMulti(string multiplied, string multiplier)
{
	int dotIndexMultiplier = multiplier.find(".");
	int dotIndexMultiplied = multiplied.find(".");
	if (dotIndexMultiplied != -1)
	{
		multiplied = multiplied.erase(dotIndexMultiplied, 1);
	}
	if ( dotIndexMultiplier !=-1){
		
		multiplier = multiplier.erase(dotIndexMultiplier, 1);
	}
	const char * multiplierString = multiplier.c_str();
	const char * multipliedString = multiplied.c_str();
	int multipliedLength = multiplied.length();
	int multiplierLength = multiplier.length();

	int realMultiplierDotIndex = multiplierLength - dotIndexMultiplier ;
	int realMultipliedDotIndex = multipliedLength - dotIndexMultiplied ;

	int realResultDotIndex = realMultipliedDotIndex + realMultiplierDotIndex;

	int sumLength = multipliedLength + multiplierLength;
	int *resultArray = new int[sumLength];
	memset(resultArray, 0, sumLength *sizeof(int));
	for (int i = 0; i < multipliedLength; i++)
		for (int j = 0; j < multiplierLength; j++)
	{
		resultArray[i + j + 1] += ((multipliedString[i] - '0')*(multiplierString[j] - '0'));
	}
	for (int resultIndex = sumLength-1; resultIndex>0; resultIndex--)
	{
		resultArray[resultIndex - 1] += resultArray[resultIndex] / 10;
		resultArray[resultIndex] = resultArray[resultIndex]% 10;
		
	}
	
	string resultString = string();

	for (int resultIndex = 0; resultIndex < sumLength; resultIndex++)
	{
		if (dotIndexMultiplied !=-1 || dotIndexMultiplier !=-1){
			if (sumLength - resultIndex == realResultDotIndex)
				resultString.push_back('.');
		}
		resultString.push_back(char('0' + resultArray[resultIndex]));
	}

	delete [] resultArray;
	return resultString;

}
string BigExponention(string base, int exp)
{
	string result = base;
	for (int i = exp; i > 1; i--)
	{
		result = BigMulti(result, base);
	}
	return result;
}
int main()
{
	//DWORD dwStart = GetTickCount(); //取windows启动到现在的流逝时间(毫秒)
	string s = string();
	int n = 0;
	while (cin >> s >> n)
	{
		string result = BigExponention(s, n);
		if (result != string("0"))
		{
		
		result.erase(0, result.find_first_not_of("0"));
		result.erase(result.find_last_not_of("0") + 1);
		}
		if (result[result.length() - 1] == '.')
		{
			result.erase(result.length()-1);
		}
		cout << result << "\n";
	}
	
	//cout << result<<"\n";
	//DWORD dwUsed = GetTickCount() - dwStart; //计算该函数所消耗的时间
	//cout << dwUsed;
}