高精度问题之大数求幂

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

.

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 and insignificant trailing zeros should be suppressed in the output.

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

解题思路: 因为做了大数相加 ，这题也就没什么好思考的。不同之处就是先去掉小数点，计算结果后在适当位置插入小数点即可。这个算法可以计算更大的数。但必须包含小数点。。。。。。

AC代码如下:

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

// 求幂  思路: 先变成整数相乘   然后根据小数的位数 结合幂  算出小数点该结果字符串的位置 即可

string bigIntegerPlus(string src,string num) {
	
	string tmp = src;
	for (int i =num.length() -1; i >= 0 ; --i) {
		
		string mid(tmp.length(),'0');
		int goBit =0;
		for (int j =  tmp.length()-1; j >= 0; --j) {
			
			int tm = goBit + (tmp[j] -'0')* (num[i] - '0');
			mid[j] = tm% 10 +'0';
			goBit = tm  /10;
		}
		
		for (int q = num.length()-1; q> i; --q) 
			mid.push_back('0');
		if (goBit != 0) 
			mid.insert(0,string(1,(char)goBit +'0'));

		// 加法运算
		if (i == num.length()-1)
			src = mid;
		else {
		
			goBit =0;
			string s(mid.length() - src.length(),'0');
			src = s + src;
			for (int j = mid.length()-1; j>=0; --j) {
			
				int tm = (mid[j] - '0') +(src[j] - '0') + goBit;
				src[j] = tm %10 + '0';
				goBit = tm /10;
			}

			if (goBit !=0) 
				src.insert(0,(char)goBit +'0'));
		}	
	}
	return src;
}

int main(int argc,char** argv) {
	
	string str;
	while ( getline(cin,str)) {
		
		// 分割出待积数 和 幂  以及小数点位置
		int i =0;
		int index = 0;// 小数位置
		int count = 0;//幂次数
		string num;
		while ( i< str.length()) {
			
			if ( str[i] != ' ') {
			
				if (str[i] == '.')
					index = i;
				else
					num.push_back(str[i]);
				++i;
				continue;
			}
			while ( !isdigit(str[i])) 
				++i;
			
			if (i + 1 == str.length())
				count = str[i] - '0';
			else 
				count = (str[i] - '0') * 10 + str[i+1] - '0';		
			break;		
		}
		
		index = num.length() - index;
			
		string res = num;
		for (int i =0; i< count-1; ++i) {
		
			res = bigIntegerPlus( res,num);
		}
		index = index * count;
		
		res.insert(res.length() - index,string(".")); 
		
		while (res.length() >1 && res[0] == '0')
			res.erase(0,1);
		
		for (int i =res.length()-1; i>=0; --i) {
			
			if (res[i] == '0' )
				res.erase(i,i+1);
			else 
				break;
		}
		cout<< res<< endl;	
	}
	
	return 0;
}