optimization

问题 Problem Given n , find the smallest positive x such that f(x) = n . f(x) is the sum of the digit sum of the factorials of the digits of x . For example, f(15) = digit_sum(1!) + digit_sum(5!) = digit_sum(1) + digit_sum(120) = (1) + (1 + 2 + 0) = 4 Breath first search can find the answer. Are there faster ways? Breath First Search def bfs(target, d_map): # Track which values of f(x) have we visited visited = set([0]) # f(x) of the current level of the search tree todo = [0] # Digits of x for

问题 I am pretty new to optimization in general and pyomo in particular, so I apologize in advance for any rookie mistakes. I have defined a simple unit commitment exercise (example 3.1 from [1]) using [2] as starting point. I got the correct result and my code runs, but I have a few questions regarding how to access stuff. import matplotlib.pyplot as plt import numpy as np import pandas as pd import shutil import sys import os.path import pyomo.environ as pyo import pyomo.gdp as gdp #necessary if

问题 Thanks to great @thatguy help I was able to create the menu dynamically, all details and explanation here: How to avoid repeating blocks of XAML in a menu It works perfectly but my problem is that at the end of the list I need to add a separator and a Delete item. This was the code I was using: <ContextMenu ItemsSource="{Binding MyTypes}" ItemContainerStyle="{StaticResource MyMenuItemStyle}"> <Separator HorizontalAlignment="Stretch" Visibility="{Binding MenuSelected.Type, Converter=

问题 Thanks to great @thatguy help I was able to create the menu dynamically, all details and explanation here: How to avoid repeating blocks of XAML in a menu It works perfectly but my problem is that at the end of the list I need to add a separator and a Delete item. This was the code I was using: <ContextMenu ItemsSource="{Binding MyTypes}" ItemContainerStyle="{StaticResource MyMenuItemStyle}"> <Separator HorizontalAlignment="Stretch" Visibility="{Binding MenuSelected.Type, Converter=

问题 I am looking for a fast modulo 10 algorithm because I need to speed up my program which does many modulo operations in cycles. I have checked out this page which compares some alternatives. As far as I understand it correctly, T3 was the fastest of all. My question is, how would x % y look like using T3 technique? I copied T3 technique here for simplicity in case the link gets down. for (int x = 0; x < max; x++) { if (y > (threshold - 1)) { y = 0; //reset total += x; } y += 1; } Regarding to

问题 I am looking for a fast modulo 10 algorithm because I need to speed up my program which does many modulo operations in cycles. I have checked out this page which compares some alternatives. As far as I understand it correctly, T3 was the fastest of all. My question is, how would x % y look like using T3 technique? I copied T3 technique here for simplicity in case the link gets down. for (int x = 0; x < max; x++) { if (y > (threshold - 1)) { y = 0; //reset total += x; } y += 1; } Regarding to

问题 I am looking for a fast modulo 10 algorithm because I need to speed up my program which does many modulo operations in cycles. I have checked out this page which compares some alternatives. As far as I understand it correctly, T3 was the fastest of all. My question is, how would x % y look like using T3 technique? I copied T3 technique here for simplicity in case the link gets down. for (int x = 0; x < max; x++) { if (y > (threshold - 1)) { y = 0; //reset total += x; } y += 1; } Regarding to

问题 I am looking for a fast modulo 10 algorithm because I need to speed up my program which does many modulo operations in cycles. I have checked out this page which compares some alternatives. As far as I understand it correctly, T3 was the fastest of all. My question is, how would x % y look like using T3 technique? I copied T3 technique here for simplicity in case the link gets down. for (int x = 0; x < max; x++) { if (y > (threshold - 1)) { y = 0; //reset total += x; } y += 1; } Regarding to

问题 I have a csv file with more than 100 columns and 3500 rows which looks like this (just an example): import pandas as pd data = pd.DataFrame(data={ 'Profit': [90, -70, 111, 40, -5, -1], 'Crit1': [True, True, False, True, False, True], 'Crit2': [False, False, False, True, True, False], 'Crit3': [True, True, False, True, True, True], 'Crit4': [False, True, True, False, False, False], 'Crit5': [True, False, False, True, True, True] }) I'd like to define 3 results: 1 - totalProfit: is the sum of

问题 This question already has answers here : How to print number with commas as thousands separators? (27 answers) Closed 7 years ago . I don't really know the "name" for this problem, so it might be a incorrect title, but the problem is simple, if I have a number for example: number = 23543 second = 68471243 I want to it make print() like this. 23,543 68,471,243 I hope this explains enough or else add comments. Any help is appreciated! 回答1: If you only need to add comma as thousand separator and