ceil | 易学教程

Math ceil()、floor()、round()方法

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Math.ceil() 功能：对一个数进行上取整。语法： Math.ceil(x) 参数： x：一个数值。返回值：返回大于或等于x，并且与之最接近的整数。注：如果x是正数，则把小数“入”；如果x是负数，则把小数“舍”。例： <script type="text/javascript"> document.write( Math.ceil(1.2)+", "+Math.ceil(1.8)+", "+Math.ceil(-1.2)+", "+Math.ceil(-1.8) ); </script> 输出结果为： document.write( Math.ceil(1.2)+", "+Math.ceil(1.8)+", "+Math.ceil(-1.2)+", "+Math.ceil(-1.8) ); 2, 2, -1, -1 Math.floor() 功能：对一个数进行下取整。语法： Math.floor(x) 参数： x：一个数值。返回值：返回小于或等于x，并且与之最接近的整数。注：如果x是正数，则把小数“舍”；如果x是负数，则把小数“入”。例： <script type="text/javascript"> document.write( Math.floor(1.2)+", "+Math.floor(1.8)+", "+Math.floor(-1.2)

round ceil floor

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转载：http://blog.csdn.net/u014624597/article/details/23868729 extern float ceilf(float); extern double ceil(double); extern long double ceill(long double); extern float floorf(float); extern double floor(double); extern long double floorl(longdouble); extern float roundf(float); extern double round(double); extern long double roundl(longdouble); round：如果参数是小数，则求本身的四舍五入。 ceil：如果参数是小数，则求最小的整数但不小于本身. floor：如果参数是小数，则求最大的整数但不大于本身. Example:如何值是3.4的话，则 3.4 -- round 3.000000 -- ceil 4.000000 -- floor 3.00000 CGRectMake(floorf(self.view.bounds.size.width*0.5f - 39.f*0.5f),self.view.bounds.size.height -57

学习PHP几个数学计算的内部函数

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简介：这是学习PHP几个数学计算的内部函数的详细页面，介绍了和php,有关的知识、技巧、经验，和一些php源码等。 class='pingjiaF' frameborder='0' src='http://biancheng.dnbcw.info/pingjia.php?id=342968' scrolling='no'> 下面主要讲述 round, floor, ceil, pow, rand,max, min, decbin, bindec, dechex, hexdec, decoct, octdec 函数。 round round - 对浮点数进行四舍五入。round 函数语法如下： round(float,precision) 其中参数 precision 表示小数点后面要保持的精度位数。如果不写参数 precision，表示四舍五入到整数位，比如： echo round(3.4); // 3echo round(3.5); // 4echo round(3.6); // 4 如果 precision 为2，表示四舍五入到小数点后2位。示例如下： echo round(1.95583, 2); // 1.96 如果参数 precision 为负数，表示四舍五入到小数点前。比如： echo round(1241757, -3); // 1242000 floor floor

Python - Ceil a datetime to next quarter of an hour

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问题 Let's imagine this datetime >>> import datetime >>> dt = datetime.datetime(2012, 10, 25, 17, 32, 16) I'd like to ceil it to the next quarter of hour, in order to get datetime.datetime(2012, 10, 25, 17, 45) I imagine something like >>> quarter = datetime.timedelta(minutes=15) >>> import math >>> ceiled_dt = math.ceil(dt / quarter) * quarter But of course this does not work 回答1: This one takes microseconds into account! import math def ceil_dt(dt): # how many secs have passed this hour nsecs =

Round up a CGFloat in Swift

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问题 How can I round up a CGFloat in Swift? I've tried ceil(CDouble(myCGFloat)) but that only works on iPad Air & iPhone 5S. When running on another simulated device I get an error saying 'NSNumber' is not a subtype of 'CGFloat' 回答1: Update : Apple have now defined some CGFloat-specific versions of common functions like ceil : func ceil(x: CGFloat) -> CGFloat ...specifically to cope with the 32/64-bit difference. If you simply use ceil with a CGFloat argument it should now work on all

Use if statement against the ones place decimal value in Excel

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问题 I need help creating a formula that rounds a number with a 1 or 6 in the ones place down to the nearest multiple of 5 (e.g., 276 to 275 or 131 to 130) and rounds any other number up to the nearest multiple of 5 (e.g., 277 to 280 or 132 to 135). I figured the logic would look something like this: =if(can't figure out this condition, ceiling(A1,5), floor(A1,5)) 回答1: You can use MROUND instead: =MROUND(A1,5) It rounds to the nearest 5. Anything including and above 277.5 will be rounded to 280,

Objective-c常用的函数

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【推荐】2019 Java 开发者跳槽指南.pdf(吐血整理) >>> 来自：http://blog.sina.com.cn/s/blog_71715bf80101bnvn.html 介绍一下Objective-c常用的函数，常数变量算术函数【算术函数】函数名说明 int rand() 随机数生成。（例） srand(time(nil));　//随机数初期化 int val = rand()P;　//0～49之间的随机数 int abs(int a) 整数的绝对值（例）int val = abs(-8); 　→8 ※浮点数的时候用fabs。 double fabs(double a) 浮点数的绝对值（例）double val = fabs(-12.345); 　→12.345 ※整数的时候用abs。 double floor(double a) 返回浮点数整数部分（舍弃小数点）（例）double val = floor(12.345); 　→12.000 double ceil(double a); 返回浮点数整数部分（舍弃小数点部分，往个位数进1）（例）double val = ceil(12.345); 　→13.000 double pow(double a, double b) a的b次方（例）double val = pow(2, 3); 　→8

Ambiguous reference to member when using ceil or round

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问题 I am just trying to use the ceil or round functions in Swift but am getting a compile time error: Ambiguous reference to member 'ceil'. I have already imported the Foundation and UIKit modules. I have tried to compile it with and without the import statements but no luck. Does anyone one have any idea what I am doing wrong? my code is as follow; import UIKit @IBDesignable class LineGraphView: GraphView { override func setMaxYAxis() { self.maxYAxis = ceil(yAxisValue.maxElement()) } } 回答1: This

Why does Binary search algorithm use floor and not ceiling - not in an half open range

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问题 When we have an array with indexes from 0 to n for example. when I use the Binary search using floor or ceiling when calculating the middle index I get the same out put. int middle = ceiling ((left+right)/2); Is there a reason using floor over ceiling ? what bug will happen using the ceiling ? 回答1: There's no difference in complexity. Both variants are log(n). Depending on the rest of your implementation, you may get a different result if your array looks for instance like 0 1 1 1 1 2 and

Why does Binary search algorithm use floor and not ceiling - not in an half open range

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