求高精度幂

周末闲着没事看了下acm的题目，要求计算高精度幂，即一个数的n次幂。 代码中的大数用STL中的string表示

本文所实现的方法效率不一定达到acm的要求(没有进行性能测试)，但是测试结果的确通过了，下面贴代码，欢迎大家拍砖。。。

源代码：

#include <string>
#include <stdio.h>

// 求两个大数相乘的结果 ( left, right 表示的数这里均为正整数 )
std::string mul( std::string left, std::string right )
{
	std::string sRet = "";
	std::string sCurr = "";
	short src = 0;
	short dst = 0;
	short ret = 0;

	// 余数
	short remainder1 = 0;
	// 商
	short quotient1 = 0;

	int counter = 0;
	for (int i = left.size()-1; i >= 0; i--)
	{
		src = left[i] - '0';
		short lastQuotient = 0;
		short remainder = 0;
		sCurr = "";
		for (int j = right.size()-1; j >= 0; j--)
		{
			dst = right[j] - '0';
			ret = src * dst;
			remainder = ret % 10;

			remainder1 = (remainder + lastQuotient) % 10;
			quotient1 = (remainder + lastQuotient) / 10;

			sCurr.insert( sCurr.begin(), '0' + remainder1 );
			lastQuotient = ret / 10 + quotient1;

			if ( j == 0 && lastQuotient > 0 )
			{
				if ( lastQuotient / 10 )
				{
					sCurr.insert( sCurr.begin(), '0' + lastQuotient%10 );
					sCurr.insert( sCurr.begin(), '0' + lastQuotient/10 );
				}
				else
				{
					sCurr.insert( sCurr.begin(), '0' + lastQuotient );
				}
			}
		}

		int tmp = counter;
		while( tmp-- )
			sCurr.push_back( '0' );

		int ndiff = sCurr.size() - sRet.size();
		for( int m = 0; m < ndiff; m++ )
			sRet.insert( sRet.begin(), '0' );

		lastQuotient = 0;
		remainder = 0;
		for (int n = sRet.size()-1; n >= 0; n--)
		{
			src = sRet[n] - '0';
			dst = sCurr[n] - '0';

			ret = src + dst;
			remainder = ret % 10;

			remainder1 = (remainder + lastQuotient) % 10;
			quotient1 = (remainder + lastQuotient) / 10;

			sRet[n] = '0' + remainder1;
			lastQuotient = ret / 10 + quotient1;

			if ( n == 0 && lastQuotient > 0 )
			{
				if ( lastQuotient / 10 )
				{
					sRet.insert( sRet.begin(), '0' + lastQuotient%10 );
					sRet.insert( sRet.begin(), '0' + lastQuotient/10 );
				}
				else
				{
					sRet.insert( sRet.begin(), '0' + lastQuotient );
				}	
			}
		}

		counter++;
	}

	return sRet;
}

// 计算大数sinput的exp次幂 （sinput表示的数和exp均大于等于0）
std::string caclExpo( std::string sinput, int exp )
{
	// 0次幂返回1
	if ( exp == 0 )
		return "1";
	
	// 移除小数点便于计算
	std::string sOutput = sinput;
	int ndot = 0;
	for ( int i = sinput.size()-1; i >= 0; i-- )
	{
		if ( sinput[i] == '.' )
		{
			ndot = sinput.size()-1-i;
			sinput.erase( i, 1 );
			break;
		}
	}
	sOutput = sinput;

	int nTotalDot = ndot;
	while( --exp )
	{
		// 顺便计算下结果中小数点后的位数
		nTotalDot += ndot;
		// 乘吧
		sOutput = mul( sOutput, sinput );
	}

	// 把小数点放到正确的位置上
	if ( nTotalDot != 0 )
		sOutput.insert( sOutput.begin() + sOutput.size()-nTotalDot, '.' );

	// 移除整数部分无效的0
	if ( sOutput.size() > 1 )
	{
		for (int i = 0; i < sOutput.size(); i++)
		{
			if ( sOutput[i] == '0' )
			{
				sOutput.erase( i, 1 );
				i--;
			}
			else
				break;
		}
	}

	// 移除小数部分无效的0
	if ( nTotalDot != 0 )
	{
		for (int i = sOutput.size()-1; i >= 0 ; i--)
		{
			if ( sOutput[i] == '0' )
			{
				sOutput.erase( i, 1 );
			}
			else
				break;
		}
	}

	return sOutput;
}

测试： 测试数据来自ACM

/* 测试数据
Sample Input
	95.123 12
	0.4321 20
	5.1234 15
	6.7592  9
	98.999 10
	1.0100 12

Sample Output
	548815620517731830194541.899025343415715973535967221869852721
	.00000005148554641076956121994511276767154838481760200726351203835429763013462401
	43992025569.928573701266488041146654993318703707511666295476720493953024
	29448126.764121021618164430206909037173276672
	90429072743629540498.107596019456651774561044010001
	1.126825030131969720661201
*/

// 测试
void Test( std::string src, int exp, std::string expected )
{
	std::string sret = caclExpo( src, exp );
	if ( sret.compare( expected ) != 0 )
		printf( "test not passed, [the result is (%s)\n, but (%s) is expected\n", sret.data(), expected.data() );
	else
		printf( "test passed\n" );
}

int _tmain(int argc, _TCHAR* argv[])
{
	Test( "95.123", 12, "548815620517731830194541.899025343415715973535967221869852721"  );
	Test( "0.4321", 20, ".00000005148554641076956121994511276767154838481760200726351203835429763013462401"  );
	Test( "5.1234", 15, "43992025569.928573701266488041146654993318703707511666295476720493953024"  );
	Test( "6.7592", 9, "29448126.764121021618164430206909037173276672"  );
	Test( "98.999", 10, "90429072743629540498.107596019456651774561044010001"  );
	Test( "1.0100", 12, "1.126825030131969720661201"  );

	return 0;
}

测试结果：