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uva748 - Exponentiation

2024-07-03 03:22:54   222人阅读

  Exponentiation 

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 $0 < n \le 25$.

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 Rn. 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

基于大数的乘法（http://blog.csdn.net/u011863942/article/details/25001689）再做处理即可，注意两点：第一，0.0的指数需要单独处理；第二，末尾的零需要删除

#include<iostream>
#include<sstream>
#include<cstdio>
#include<string>
#include<cstring>
using namespace std;

#define MAX 1000000
int ans[MAX];

string Multi(string str1, string str2){
	int i;
	stringstream s;
	string str;

	memset(ans, 0, sizeof(ans));
	for(i = str1.size()-1; i >= 0; i--){
		for(int j = str2.size()-1, p = str1.size()-1-i; j >=0; j--, p++){
			ans[p] += (str1[i]-‘0‘)*(str2[j]-‘0‘);
		}
	}

	for(i = 0; i < MAX; i++){
		ans[i+1] += ans[i]/10;
		ans[i] = ans[i]%10;
	}

	for(i = MAX-1; i >=0; i--){
		if(ans[i])
			break;
	}
	if(i == -1)
		return "0";
	else{
		for(int j = i; j >= 0; j--){
			s << ans[j];
		}
		s >> str;
		return str;
	}	
}

int main(void){
	string str, ans;
	int n;

#ifndef ONLINE_JUDGE
	freopen("f:\\infile.txt", "r", stdin);
#endif

	while(cin >> str >> n){
		int dot, ansdot, i;
		ans = "1";
		dot = str.find(‘.‘, 0);
		ansdot = (str.size()-1-dot)*n;
		str.erase(dot, 1);
		for(i = 0; i < n; i++){
			ans = Multi(ans, str);
		}
		if(ans == "0"){
			cout << ".0" << endl;
			continue;
		}
		else{
			if(ans.size() < ansdot)
				ans.insert(0, ansdot-ans.size(), ‘0‘);
			ans.insert(ans.size()-ansdot, 1, ‘.‘);
		}
		for(i = ans.size()-1; i >=0; i--){
			if(ans[i] != ‘0‘)
				break;
		}
		ans.erase(ans.begin()+i+1, ans.end());
		cout << ans << endl;
	}
	return 0;
}


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