floor

So in c++11 the Chrono Library provides, duration_cast : Computations are done in the widest type available and converted, as if by static_cast, to the result type only when finished And c++17 's floor : Returns the greatest duration t representable in ToDuration that is less or equal to d So for all x will the result of these 2 calls be equal: chrono::duration_cast<chrono::seconds>(x) chrono::floor<chrono::seconds>(x) As far as I can tell, same as the difference between static_cast and std::floor : Negatives are rounded down instead of truncated toward zero. #include <iostream> #include

I'm just learning haskell (on my own, for fun) and I've come up against a wall. My Question: How can I define a function flrt = (floor . sqrt) When I try it in a file and compile, GCHi complains with the following: AKS.hs:11:9: No instance for (RealFrac Integer) arising from a use of `floor' Possible fix: add an instance declaration for (RealFrac Integer) In the first argument of `(.)', namely `floor' In the expression: (floor . sqrt) In an equation for `flrt': flrt = (floor . sqrt) AKS.hs:11:17: No instance for (Floating Integer) arising from a use of `sqrt' Possible fix: add an instance

B树 tips：floor向上取整 B树的定义 B树，又称多路平衡查找树， B树中所有结点孩子结点数的最大值称为B树的阶 ,通常用m表示。 一棵m阶B树或者空树，或为满足以下条件的m叉树： 树中每个结点最多有m棵子树（即最多有m-1个关键字） 若根节点不是终端结点，那么最少有两颗子树 除根节点之外的所有非叶结点至少有floor[m/2]棵子树（即最少有floor[m/2] - 1个关键字） B树的高度 若n≥1，则对任意一棵包含n个关键字，高度为h，阶数为m的B树 因为B树每个结点最多有m 棵子树，m-1个关键字，所以在一棵高度为h的m阶B树中关键字的个数应该满足 ，因此有 ​ 若让每个结点中的关键字个数达到最少，则容纳同样多关键字的B树高度达到最大。由B树的定义可知，第一层至少有1个结点；第二层至少有2个结点，除了根节点之外的所有非叶结点至少有floor[m/2]棵子树，第三层至少有2*floor[m/2]个节点，以此类推，注意到第h+1层节点是不包含任何信息的叶节点。对于关键字个数为n的B树，叶节点即查找不成功的节点为n+1，因此有 ,即 ​ 来源： https://www.cnblogs.com/masterchd/p/11566882.html

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

The differences reside in the returned value giving inputs around tie-breaking I believe, such as this code : int main() { std::cout.precision(100); double input = std::nextafter(0.05, 0.0) / 0.1; double x1 = floor(0.5 + input); double x2 = round(input); std::cout << x1 << std::endl; std::cout << x2 << std::endl; } which outputs: 1 0 But they are just different results in the end, one chooses its preferred one. I see lots of "old" C/C++ programs using floor(0.5 + input) instead of round(input) . Is there any historic reason? Cheapest on the CPU? haccks std::round is introduced in C++11. Before

1 #encoding:utf-8 2 import math 3 4 #向上取整 5 print "math.ceil---" 6 print "math.ceil(2.3) => ", math.ceil(2.3) 7 print "math.ceil(2.6) => ", math.ceil(2.6) 8 9 #向下取整 10 print "\nmath.floor---" 11 print "math.floor(2.3) => ", math.floor(2.3) 12 print "math.floor(2.6) => ", math.floor(2.6) 13 14 #四舍五入 15 print "\nround---" 16 print "round(2.3) => ", round(2.3) 17 print "round(2.6) => ", round(2.6) 18 19 #这三个的返回结果都是浮点型 20 print "\n\nNOTE:every result is type of float" 21 print "math.ceil(2) => ", math.ceil(2) 22 print "math.floor(2) => ", math.floor(2) 23 print "round(2) => ", round(2) 运行结果： 来源：

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