习题3-12(uva-11809)

最新推荐文章于 2022-11-07 16:59:41 发布

原创最新推荐文章于 2022-11-07 16:59:41 发布 · 105 阅读

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本文通过具体的程序实现，探讨了如何将浮点数转换为特定形式的指数表示法。通过对不同范围内的浮点数进行分解，计算其底数和指数部分，并采用查找表的方式匹配最接近的指数表示。

#include <iostream>
#include <vector>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <cstring>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <numeric>
#include <chrono>
#include <ctime>
#include <cmath>
#include <cctype>
#include <string>
#include <cstdio>
#include <iomanip>


#include <thread>
#include <mutex>
#include <condition_variable>
#include <functional>
#include <iterator>
using namespace std;
const int RLEN = 10, CLEN = 32,MAXSIZE = 1e5;

double M[RLEN][CLEN] = { 0 };
long long int   E[RLEN][CLEN] = { 0 };
char input[MAXSIZE] = { 0 };


int main()
{
	double temp = 0;
	/*
	m * 2^ e = A e B
	log(m*2^e) = log(A) + B
	log(m) + elog(2) = log(A) + B
	
	*/
	for (int i = 0; i <= 9; ++i) {
		for (int j = 1; j <= 30; ++j) {
			double m = 1 - pow(2, -(i + 1)),e = pow(2,j)-1;
			double t = log10(m) + e * log10(2);
			E[i][j] = t;
			M[i][j] = pow(10,t - E[i][j]);
		}
	}

	while (scanf("%s",input) && strncmp(input,"0e0",4) != 0)
	{
		int ntime = 0,index = 0,len = strlen(input);
		double base = 0;
		while (index < len && input[index] != 'e') ++index;
		for (int i = index + 1; i < len; ++i)
			ntime = ntime * 10 + (input[i] - '0');
		input[index] = '\0';
		base = atof(input);

		int x = 0, y = 0;
		for (int i = 0; i <= 9; ++i) {
			for (int j = 0; j <= 30; ++j) {
				if (E[i][j] == ntime && fabs(M[i][j] - base) < 1e-4) {
					x = i; y = j;
					i = 10; break;
				}
			}
		}
		cout << x << " " << y << endl;
	}
	
	return 0;
}