broqdev

#include "string.h"

void reverse_cstr(char* cstr);
void reverse_cstr_complex(char* cstr);

static void reverse_char(char* cstr, int len, int reverse_count);
static void reverse_inner(int* left, int* right);
static void reverse_int(char* cstr, int len, int reverse_count);

void test() {

broqdev

class Foo(object):
    def __init__(self):
        self._func = None

    def callDerivedFunc(self):
        # self._func() # call bound method
        # self._func(self) # call unbound method
        method = self._func
        method(self) # call unbound method

broqdev

arg_count = foo.func_code.co_argcount
args = foo.func_code.co_varnames[:arg_count]

args_dict = {}
for k, v in d.iteritems():
    if k in args:
        args_dict[k] = v

foo(**args_dict)

broqdev

--- install2.js  2013-01-25 15:32:38.358227000 +0800
+++ install.js	2013-01-25 14:29:55.639011700 +0800
@@ -108,7 +108,9 @@
     unzipStream.on('error', finishIt)
     unzipStream.on('close', finishIt)
 
-    var readStream = fs.createReadStream(downloadedFile)
+    var zipPath = path.normalize('d:/node/phantomjs-1.8.0-windows.zip')
+    var readStream = fs.createReadStream(zipPath)
+    //var readStream = fs.createReadStream(downloadedFile)

broqdev

# -*- coding: utf-8 -*- 

'''
A circular array A[1, . . . , n] is an array such that n ≥ 3 and A[1] and A[n] are considered to be adjacent elements just like A[1] and A[2], A[2] and A[3], etc., are adjacent.
We are given a circular array A[1, . . . , n] of non-negative integers. For any i, j ∈ {1, 2, . . . , n} such that i != j, dist(i, j) = max {|i − j|, n − |i − j|}.
Design a linear time algorithm that computes a maximum number t such that for some i, j ∈ {1, 2, . . . , n}, i != j, t = A[i] + A[j] + dist(i, j).
'''

def distant(i, j, n):
    dis = abs(j-i)

broqdev

def distant(i, j, n):
    i%=n
    j%=n
    dis = abs(j-i)

    if ((n-dis) > dis):
        dis = n-dis

    return dis

broqdev

# deno = [2, 5, 13]
# deno_nums = [1, 3, 1]
# target = 17

deno = [1, 2, 7, 11]
deno_nums = [5, 18, 1, 7]
target = 82

# every cell has 4 parts:
# one pointer to the previous cell which is part of current combination

broqdev

当前是i，j in [0...i)，first pass算max(A[i]+A[j]+i-j)，second pass 算max(A[i]+A[j]+n-(i-j))

第一次用公式max_value = max(max_value+1, A[i-1]+1)

第二次用公式max_value = max(max_value-1, A[i-1]+n-1)

for i in xrange(len(ar)):
     #try to update max_value
     ret = max(ret, A[i]+max_value)
这个循环写两次，用两个不同的公式

broqdev

Installing cx_Oracle (x86_64) for Mac OSX (tested on El Capitan 10.11.4)

Assume you've got homebrew installed.

Assume python is installed through

brew install python --universial

Install instant client

broqdev

package com.test

import java.net.{URI, URL}

import org.apache.hadoop.conf.Configuration
import org.apache.hadoop.fs.FsUrlStreamHandlerFactory
import org.apache.spark.sql.SparkSession
import org.apache.hadoop.fs.Path;

// build into a jar and run with spark-submit