[POJ] #1003# Hangover : 浮点数运算】的更多相关文章

一. 题目 Hangover Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 116593   Accepted: 56886 Description How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (W…
POJ.1003 Hangover ( 水 ) 代码总览 #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <vector> #define nmax using namespace std; vector <double> v…
链接地址: Poj:http://poj.org/problem?id=1003 OpenJudge:http://bailian.openjudge.cn/practice/1003 题目: Hangover Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 94993   Accepted: 46025 Description How far can you make a stack of cards overhang…
    Hangover Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 95164   Accepted: 46128 Description How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We'r…
Hangover Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 99450   Accepted: 48213 Description How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're as…
Hangover Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 103896   Accepted: 50542 Description How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're a…
#include <iostream> using namespace std; int main() { double len; while(cin >> len && len) { double sum = 0.0; double i = 1.0; ; while(sum < len) { sum += i/n; ++n; } cout << n- << " card(s)" <<endl; } ;…
一. 题目 Financial Management Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 173910   Accepted: 65186 Description Larry graduated this year and finally has a job. He's making a lot of money, but somehow never seems to have enough. Larry ha…
js,java浮点数运算错误及应对方法 一,浮点数为什么会有运算错误 IEEE 754 标准规定了计算机程序设计环境中的二进制和十进制的浮点数自述的交换.算术格式以及方法. 现有存储介质都是2进制.2进制的进制基数是2,那么一个数字只要被因素包含大于2的质数的数除,都会产生无限循环小数.无限循环小数和无理数都无法,和非无限循环的有理数一起用同一种方式存储到存储介质上的同时还保持计算的兼容性. 对于无限循环小数,可以设计一种格式存储到介质上,但是同时又要和非无限循环的有理数能够计算,效率应该会变得…
前言 上一章我们简单介绍了IEEE浮点标准,本次我们主要讲解一下浮点运算舍入的问题,以及简单的介绍浮点数的运算. 之前我们已经提到过,有很多小数是二进制浮点数无法准确表示的,因此就难免会遇到舍入的问题.这一点其实在我们平时的计算当中会经常出现,就比如之前我们提到过的0.3,它就是无法用浮点小数准确表示的. 为此LZ专门写了一个小程序,使用Java语言打印出了0.3的二进制表示,是这样的一个数字,0 01111101 00110011001100110011010.我们来简单算一下,这个数值大约是…