position

Eight queens经典回溯算法

本秂侑毒 提交于 2019-12-11 14:39:49
def conflict(state, nextX): print('state:',state) nextY = len(state) for i in range(nextY): print("i=",i) if abs(state[i]-nextX) in (0, nextY-i): print('confict') return True return False def queens(num=8, state=()): for position in range(num): print('***') print('pos:',position) if not conflict(state, position): if len(state) == num-1: print('back to 1st_space_____________position:',position) yield (position,) else: print('Into 2nd_space__________recursive') for result in queens(num, state + (position,)): print('membersihip position:',position) print('membership result:',result) yield

Java - Add Element to an Array

怎甘沉沦 提交于 2019-12-11 13:55:26
问题 Looked around, couldn't find any similar questions in java.. Basically I need to add a number to an int array in a specific position index I can only use Arrays, no ArrayLists Here is what I have so far, and I know why it doesn't work, but I can't figure out how to fix that problem of overwriting, which I don't want it to do. The task is a non-overwriting insert. e.g. the final result would be [1 2 1337 3 4 5 6 7 8] Here is the code snippet: public void main(String[] args) { int[] array = {1

List a specific position in each list within a list (python)

有些话、适合烂在心里 提交于 2019-12-11 13:18:08
问题 Is there a way to select every 2nd or 3rd (for example) item within a matrix? For example: f = [["1", "5", "8", "9"], ["2", "6", "9", "10"], ["3", "7", "11", "12"]] I am wondering if there is a direct function to select every 2nd number in every list (preferably putting those digits in a list as well). Thus resulting into: ["5", "6", "7"] I know that I can achieve this using a loop but I am wondering if I can achieve this directly. 回答1: Without any loop (external) >>> f = [["1", "5", "8", "9"

炫酷的CSS3响应式表单

五迷三道 提交于 2019-12-11 12:52:23
原创YouTube@ Online Tutorials css代码: * { margin: 0; padding: 0; box-sizing: border-box; font-family: 'Poppins', sans-serif; } body { display: flex; justify-content: center; align-items: center; min-height: 100vh; background: #03080a; } .container { width: 80%; padding: 20px; } .container h2 { width: 100%; color: #45f3ff; font-size: 36px; text-align: center; margin-bottom: 10px; } .container .row100 { position: relative; width: 100%; display: grid; grid-template-columns: repeat(auto-fit, minmax(300px, 1fr)); } .container .row100 .col { position: relative; width: 100%; padding: 0 10px; margin:

聚宽源码45

大兔子大兔子 提交于 2019-12-11 12:37:24
原文策略源码如下: #HS300–随机森林拐点识别 import math import numpy as np import pandas as pd from pandas import DataFrame,Series import pickle import time import datetime import jqdata import talib as tb import sklearn from sklearn.decomposition import PCA from sklearn.tree import DecisionTreeRegressor from sklearn.ensemble import AdaBoostClassifier,GradientBoostingClassifier,RandomForestRegressor,RandomForestClassifier from sklearn import cross_validation, metrics,svm,preprocessing from sklearn.metrics import mean_squared_error from sklearn.cross_validation import train_test_split from sklearn.grid_search

Sticky footer acting weird

萝らか妹 提交于 2019-12-11 11:41:56
问题 I have made a sticky footer for my website, but it doesnt act like i want i to. I want it to be totally at the bottom, always depending on the length of the content. Instead it is always present on screen, as it is. Can anyone help me ? This site footer css: #footer{ position: fixed; bottom: 0; z-index: 1; width: 100%; height: 50px; clear: both; background-image: url("../images/footer_pattern.png"); } 回答1: #footer{ width: 100%; height: 50px; clear: both; background-image: url("../images

Imagick annotateImage: how to set text position from the top left

懵懂的女人 提交于 2019-12-11 11:17:52
问题 I'm trying to create a wrapper function around annotateImage to be able to set the exact top and left positions of a given text. The default method sets the y-position from the baseline, which means there has to be a lot of experimentation involved if one wants to draw a text at an exact spot on an image. This is what I mean... $image->annotateImage($draw, 0, 0, 0, 'The quick brown fox'); In the above code, the text is invisible because the y position is 0. So to fix this, I've started with

THREE.JS, ignore parent's rotation

邮差的信 提交于 2019-12-11 11:08:44
问题 I'm trying to make child object follow parent's position and behave like a normal child, however I want it to keep it's rotation unchanged. What's the best way to do that without affecting performance(I'm tight on cpu budget already running 2 workers and having lots of objects)? Is there a setting which would allow to only child's position being affected? Also an important thing is that when parent is being rotated then child's position should follow that rotation. 回答1: You want to prevent a

android-gallery how to set the position of the child which is selected

隐身守侯 提交于 2019-12-11 10:53:53
问题 as we know,the default position of the child which is selected is in the middle.but i want to show it in the gallery's left.how to do this. thanks~ 回答1: use the method setSelection(int position) to set which item will be shown as selected. For showing the left you have just use setSelection(0). simple... please check the given link by Rahul for more information about Gallery 来源: https://stackoverflow.com/questions/4710413/android-gallery-how-to-set-the-position-of-the-child-which-is-selected

JS pageX and pageY take the coordinate plus the margin of relative position div?

北慕城南 提交于 2019-12-11 10:49:36
问题 I have a div that I want to center so I use margin-left: auto; margin-right: auto; position: relative; That's the easiest way to center I admit, but when I want to use event.pageX and event.pageY it takes the coordinates plus the left margin and that's wrong. Here is the fiddle. Click somewhere on the green rectangle to watch the result : http://jsfiddle.net/FGkUq/ Any ideas how to fix so to show the square to the coordinates without the left margin ? 回答1: Take a look at the updated fiddle.