ceil

整数 整数没有上下限 >>> 27**100 136891479058588375991326027382088315966463695625337436471480190078368997177499076593800206155688941388250484440597994042813512732765695774566001 运算符 “//” 整除，会舍掉小数 >>> 5/3 1.6666666666666667 >>> 5//3 1 >>> 1/3 0.3333333333333333 >>> 1//3 0 浮点数 数值大小有上下限 >>> 500 ** 10000

1. Round （四舍六入五留双） 功能说明：对一个实数进行四舍五入。(按照银行家算法) 例： var i, j: Integer; begin i := Round(1.5); // i等于2 j := Round(2.5); // j等于2 end; 在 Delphi 中使用Round函数得到的答案有时与我们所预期的会不太一样：采用的是四舍六入五留双。即当舍或入位大于或小于五时按四舍五入来处理 ，而当舍或入位等于五时，就要看前面一位是什么，根据奇进偶不进，它总是返回一个偶数值。 例： i:= Round(11.5)//i等于12 i:= Round(10.5)//i等于10 这种Round其实是按照银行家算法，统计学上一般都用这种算法，比传统的"四舍五入"要科学。 如果要使用传统的"四舍五入"方法，可以使用下面函数： function RoundClassic(R: Real) 2. trunc （取得X的整数部分） 如： trunc (-123.55)=-123, floor(123.55)=123 3. ceil （取得大于等于X的最小的整数） 如：ceil(-123.55)=-123， ceil(123.15)=124 4. floor （取得小于等于X的最大的整数） 如：floor(-123.55)=-124，floor(123.55)=123 5.RoundTo

Do you know how to do this simple line of code without error using Boost::multiprecison ? boost::multiprecision::cpp_int v, uMax, candidate; //... v += 6 * ceil((sqrt(uMax * uMax - candidate) - v) / 6); Using MSVC there is an error for "sqrt" and it's possible to fix it with: v += 6 * ceil((sqrt(static_cast<boost::multiprecision::cpp_int>(uMax * uMax - candidate)) - v) / 6); Then there is an error for "ceil" and it's possible to fix it with: namespace bmp = boost::multiprecision; typedef bmp::number<bmp::cpp_dec_float<0>> float_bmp; v += 6 * ceil(static_cast<float_bmp>((sqrt(static_cast<bmp:

For an arbitrary value 'v' of a floating point type (float/double/long double), does C89 guarantee that the mathematically exact integer result of floor(v) and ceil(v) is a representable value of the type of 'v'? Does any of the later C or C++ standards guarantee this? Does IEEE 754 guarantee this? This is guaranteed by the construction of IEEE-754 numbers. (To be clear: C does not guarantee IEEE-754, but the following analysis holds for all other floating-point formats with which I am familiar as well; the crucial property is that all sufficiently large numbers in the format are integers).

问题描述： 将数值处理为 10 的倍数，并支持向上或者向下取整 如将 2345 可以处理为 2300 | 2400 | 3000 | 2000 解决方案： /** * 将数字取整为10的倍数 * @param {Number} num 需要取整的值 * @param {Boolean} ceil 是否向上取整 * @param {Number} prec 保留多少位（从左到右） */ const formatInt = (num, ceil = true, prec = 2) => { const len = String(num).length; if (len < prec) { return }; const mult = Math.pow(10, (len - prec)); return ceil ? Math.ceil(num / mult) * mult : Math.floor(num / mult) * mult; } // formatInt(2345, true, 2) -> 2400 // formatInt(2345, false, 2) -> 2300 // formatInt(2345, true, 1) -> 3000 // formatInt(2345, false, 1) -> 2000 来源： https://www.cnblogs.com

I know that C++ provides us with a ceil function. For practice, I was wondering how can we implement the ceil function in C++. The signature of the method is public static int ceil(float num) Please provide some insight. I thought of a simple way: Convert num to a string, find the index of the decimal point, check if the decimal part is greater than 0. If yes, return num+1 else return num. But I want to avoid using the string conversion Here is a naive implementation for positive numbers (this uses the fact that casting to (int) truncates toward zero): int ceil(float num) { int inum = (int)num

I have a pandas series series . If I want to get the element-wise floor or ceiling, is there a built in method or do I have to write the function and use apply? I ask because the data is big so I appreciate efficiency. Also this question has not been asked with respect to the Pandas package. You can use NumPy's built in methods to do this: np.ceil(series) or np.floor(series) . Both return a Series object (not an array) so the index information is preserved. You could do something like this using NumPy's floor, for instance, with a dataframe : floored_data = data.apply(np.floor) Can't test it

ceil向上取整 floor向下取整 ceil的实例 #include <iostream> #include <math.h> using namespace std; int main() { double a[10]; for (int n=0;n<10;n++) { cout<<"输入a："<<endl; cin>>a[n]; a[n]=ceil(1.0*a[n]/8); cout<<"运算结果：\na="<<a[n]<<endl; } system("pause"); return 0; } 输出结果： floor的实例 #include <iostream> #include <math.h> using namespace std; int main() { double b[11]; cout<<"输入b："<<endl; for (int i=0;i<10;i++) { cin>>b[i]; b[i]=floor(1.0*b[i]/8); cout<<"b="<<b[i]<<endl; } system("pause"); return 0; } 输出结果： 来源： CSDN 作者： 胡大锤锤 链接： https://blog.csdn.net/qq_38834877/article/details/103239616

Possible Duplicate: PHP Round function - round up to 2 dp? What my problem is: When i use ceil(3.6451895227869); i get like 4 but i want 3.65 Can you help me out? UPDATE Please remember: This should always round to ceil like while rounding 3.6333333333333 it must not be 3.63 but should be 3.64 Check out http://www.php.net/manual/en/function.round.php <?php echo round(3.6451895227869, 2); ?> EDIT Try using this custom function http://www.php.net/manual/en/function.round.php#102641 <?php function round_up ( $value, $precision ) { $pow = pow ( 10, $precision ); return ( ceil ( $pow * $value ) +