[poj] 1001 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 R ⁿ 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

分析：肯定不是不能用double和long double了，因为在相乘时会损失精度。所以用string类型来模拟double的乘法和加法，只要按照手算的方法来实现就可以了（有两个细节，刚开始我把数组反转了，后来又把小数点去掉，这样利于逻辑思维，当然也可以不用这用）

#include<iostream>
#include<fstream>
#include<vector>
#include<string>
#include<algorithm>
#include<cmath>
#include<string.h>
using namespace std;

//单个位相乘，累加到index开始的结果中
void douMulOne(string &x,int index,char ch1,char ch2)
{
	while(index>x.size())
	{//把0加上
		x.push_back('0');
	}
	int one=ch1-'0';
	one =one*(ch2-'0');
	while(1)
	{
		if(one==0)
			break;
		if(index<x.size())
		{
			one=one+(x[index]-'0');
			x[index]=one%10+'0';
		}
		else
		{
			x.push_back(one%10+'0');
		}
		one=one/10;
		index++;
	}
}

string douMul(string x1,string x2)
{
	//去掉小数点，简化逻辑
	string::size_type count1 = x1.find('.');
	string::size_type count2 = x2.find('.');
	count1>=x1.size()?count1=0:count1=count1;
	count2>=x2.size()?count2=0:count2=count2;

	if(count1)
		x1.erase(x1.begin()+count1);
	if(count2)
		x2.erase(x2.begin()+count2);

	string result("0");
	for(string::size_type i=0;i<x2.size();i++)
	{
		for(string::size_type j=0;j<x1.size();j++)
		{
			douMulOne(result,i+j,x2[i],x1[j]);
		}//end for
	}//end for
	while(count1+count2>=result.size())
		result.push_back('0');

	if(count1+count2)//还原小数点
		result.insert(result.begin()+count1+count2,'.');
	return result;
}

void print(string sum)
{
	while(1)
	{
		//按照格式输出
		string::size_type count = sum.find('.');
		count>=sum.size()?count=0:count=count;
		if((sum[0]=='0'&&count>0)||sum[0]=='.')
			sum.erase(sum.begin());
		else if(sum[sum.size()-1]=='0')
			sum.erase(sum.end()-1);
		else break;
	}
	reverse(sum.begin(),sum.end());
	cout<<sum<<endl;

}

int main()
{
	string r;
	int n;
	while(cin>>r>>n)
	{
		string sum("1");
		//反转，逻辑简单
		reverse(r.begin(),r.end());
		while(n--)
			sum=douMul(sum,r);
		print(sum);
	}//end while
}