POJ: Exponentiation

Description

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.

Input

The input will consist of a set of pairs of values for R and n. The R value will occupy columns 1 through 6, and the n value will be in columns 8 and 9.

Output

The output will consist of one line for each line of input giving the exact value of R^n. Leading zeros should be suppressed in the output. Insignificant trailing zeros must not be printed. Don't print the decimal point if the result is an integer.

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

Hint

If you don't know how to determine wheather encounted the end of input: 

s is a string and n is an integer 

代碼：

/*1001: Exponentiation
先不考慮小數，問題就轉化爲整數的冪，可以使用大數相乘來解決。
然後再對結果加入小數點。
*/

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

string BigNumMultipy( string str1, string str2 );

int main()
{
	string input;
	while( getline( cin, input ) )
	{
		if( input.length() != 9 )
		{
			cout << "" << endl;
			continue;
		}
		string zhenshu = "";
		int n = 0;
		int dotCount = 0;
		int index = input.find( "." );
		if( index == string::npos )
		{
			dotCount = 0;
			zhenshu = input.substr( 0, 6 );
			n = atoi( input.substr( 7, 2 ).c_str() );
		}
		else
		{
			dotCount = 5 - index;
			//if( atoi( input.substr( index + 1, dotCount ).c_str() ) != 0 )
				zhenshu = input.substr( 0, index ) + input.substr( index + 1, dotCount );
			//else
			//	zhenshu = input.substr( 0, index );
			n = atoi( input.substr( 7, 2 ).c_str() );
		}
		//cout << zhenshu << ' ' << n << ' ' << dotCount << endl;
		//循環大數相乘
		string res = zhenshu;
		if( atoi( zhenshu.c_str() ) == 0 )
		{
			cout << "0" << endl;
			continue;
		}
		int i, j, k;  //poj不允許在for在初始化
		for( i = 1; i < n; i++ )
		{
			res = BigNumMultipy( res, zhenshu );
		}
		//cout << res << endl;
		int totalDotCount = dotCount * n;
		if( totalDotCount != 0)  //只有小數才輸出'.'
			res = res.substr( 0, res.length() - totalDotCount ) + "." + res.substr( res.length() - totalDotCount, totalDotCount );

		for( j = 0; j < res.length() && res[j] == '0'; j++ );  //前導0不能輸出
		for( k = res.length() - 1; k >= 0 && res[k] == '0' && dotCount != 0; k-- );  //小數點後的無效0也不輸出
		if( res[k] == '.' )
			k--;

		res = res.substr( j, k + 1 - j );
		cout << res << endl;
	}
	return 0;
}


//數組實現大數相乘
string BigNumMultipy( string str1, string str2 )
{
	int len1 = str1.length();
	int len2 = str2.length();
	vector< int > res( len1 + len2, 0 );
	reverse( str1.begin(), str1.end() );
	reverse( str2.begin(), str2.end() );
	int i, j, k;
	for( i = 0; i < len1; i++ )
	{
		for( j = 0; j < len2; j++ )
		{
			res[i+j] += ( str1[i] - '0' ) * ( str2[j] - '0' );
		}
	}
	for( k = 0; k < len1 + len2; k++ )
	{
		if( res[k] >= 10 )
		{
			res[k+1] += res[k] / 10;  //這兩句順序一定要正確，先算進位，然後纔是本位。
			res[k] %= 10;
		}
	}
	string resStr = "";
	for( k = 0; k < len1 + len2; k++ )
	{
		char c = '0' + res[k];
		resStr += c;
	}

	reverse( resStr.begin(), resStr.end() );
	return resStr;
}

1) 先把小數轉化爲整數，問題就轉化爲大數的冪，可以用循環調用大數相乘解決，然後再對結果做小數處理；

2）大數相乘的模塊要熟練；

3）需要考慮一些特殊情況，例如：輸入整數則結果不能有小數點、前導零不能輸出、小數點後的無效零也不能輸出；