嗨,
也许是因为它是星期五,但我无法理解这一点:
(Double)1/2 = 0.5
(Double)1/(Double)2 = 0.5
(Double)((Double)1/(Double)2) = 0.5
(Double)(1/2) = 0.0
为什么最后一次操作是0？:S
亲切的问候.
7/2 = 3.5
如何获得大量余数？在这个例子中它应该是4而不是3.
我在div中有两个输入,我希望将它们分开.
<div>
<input type="number" id="a"> / <input type="number" id="b">
<input type="submit">
<p class="result">RESULT HERE</p>
</div>
如何用jquery完成这个数学运算？
我有一个元组列表
b = [('676010', '5036043'), ('771968', '4754525'), ('772025', '4754525'), ('772072', '4754527'), ('772205', '4754539'), ('772276', '4754542'), ('772323', '4754541'), ('647206', '5036049')]
我正在使用map()它们将它们转换为类型float.
In [45]: print [map(float,e) for e in b]
[[676010.0, 5036043.0], [771968.0, 4754525.0], [772025.0, 4754525.0], [772072.0, 4754527.0], [772205.0, 4754539.0], [772276.0, 4754542.0], [772323.0, 4754541.0], [647206.0, 5036049.0]]
如何对元素应用数学运算map(),比方说除以100,000？
我知道以不同的方式到达那里,但我如何实现这一目标map()？
In [46]: print [(float(tup[0])/100000,float(tup[1])/100000) for tup in b]
[(6.7601, 50.36043), (7.71968, 47.54525), (7.72025, 47.54525), (7.72072, 47.54527), (7.72205, 47.54539), (7.72276, 47.54542), (7.72323, 47.54541), (6.47206, 50.36049)]
我能否始终相信,当我划分两个精确代表整数的IEEE 754双精度数时,我会得到确切的值？
例如:
80.0/4.0
=> 20.0
结果是完全正确的20.0.
是否存在一些结果不准确的潜在价值19.99999999？
现在这可能是我在这里非常愚蠢,但请看下面的代码.
我正在尝试计算出我运行脚本时我已经消耗的碳水化合物目标的百分比.我得到总数并存储在carbsConsumed和carbsGoal.carbsPercent然后计算消耗的百分比.但是,carbsPercent每次返回0.有什么想法吗？
#!/usr/bin/env python2.7
import myfitnesspal
from datetime import datetime
username = 'someusername'
password = 'somepassword'
date = datetime.now()
client = myfitnesspal.Client(username, password)
day = client.get_date(date.year, date.month, date.day)
#day = client.get_date(2015,11,12)
carbs = 'carbohydrates'
carbsConsumed = day.totals[carbs]
carbsGoal = day.goals[carbs]
carbsPercent = (carbsConsumed / carbsGoal) * 100
print 'Carbs consumed: ' + str(carbsConsumed)
print 'Carbs goal: ' + str(carbsGoal)
print 'Percentage consumed: ' + str(carbsPercent)
我计算的平均分如下:
average_score=$(awk "BEGIN {printf \"%.2f\",${sum_of_score}/${number_of_lines}}")
其中sum_of_scores按每个greped ID计算如下:
sum_of_score=$(grep 271712:E1 M10.6.txt | awk '{s+=$5} END {print s}')
number_of_lines=$(grep 271712:E1 M10.6.txt | awk 'END{print FNR}')
但是有时sum_of_score和/或number_of_lines的值可能为零,因此我收到错误:
awk: BEGIN {printf "%.2f",/0}
awk: ^ unterminated regexp
awk: cmd. line:1: BEGIN {printf "%.2f",/0}
awk: cmd. line:1: ^ unexpected newline or end of string
我该如何处理这个错误？
我对此问题有些困惑：
int a = 5, b = 2;
double c = a / b;
cout << c;
输出：
2
为什么呢
我可以通过使用以下方式通过：
double aa = a,
bb = b;
c = aa / bb;
输出：
2.5
救命 ！:(
我有以下代码片段：
use num_bigint::*; // 0.2.2
use num_traits::*; // 0.2.8
use std::ops::*;
fn xgcd(b: &BigInt, a: &BigInt) -> (BigInt, BigInt, BigInt) {
let mut x0: BigInt = One::one();
let mut x1: BigInt = Zero::zero();
let mut y0: BigInt = Zero::zero();
let mut y1: BigInt = One::one();
let mut q = 0;
while a != Zero::zero() {
q = b.div(a);
b = a;
a = &b.rem(a);
x0 = x1;
x1 = x0 - q * x1;
y0 = y1;
y1 = y0 …