List of Comments

As in put your chr file on screen

Hey dude how would you do that with chr files for your character?

I know, this just demonstrates the basis, the prototyping.

You could also use some assertion libraries like 'should', 'expect' and 'assert'. For unit test, you can use Chai or Mocha.

how i want to move code using phyton to c++.....please help me......this is my code using phyton....and i wan to move from phyton to c++... print "Welcome To The Resistor Calculator" print "Our Group Members\nMuhammad Harith Bin Kamlyzan\nMuhammad Hisyam Bin Harmizon\nIkhwan Hakim Bin Mohd Asri\nShahrul Izwan Bin Kamal Akhball\nMuhammad Shahzlan Bin Mohd Nizam" start = raw_input("Press Enter To Start") first_colours = {"Black","Brown","Red","Orange","Yellow","Green","Blue","Purple","Grey","White"} while True: first_colour = raw_input("\nPlease State The First Colour Band\nBlack\nBrown\nRed\nOrange\nYellow\nGreen\nBlue\nPurple\nGrey\nWhite\nYour Choice?\n") if first_colour == "Black": first_colour = 0 break elif first_colour == "Brown": first_colour = 1 break elif first_colour == "Red": first_colour = 2 break elif first_colour == "Orange": first_colour = 3 break elif first_colour == "Yellow": first_colour = 4 break elif first_colour == "Green": first_colour = 5 break elif first_colour == "Blue": first_colour = 6 break elif first_colour == "Purple": first_colour = 7 break elif first_colour == "Grey": first_colour = 8 break elif first_colour == "White": first_colour = 9 break else: print "Invalid colour name" second_colours = {"Black","Brown","Red","Orange","Yellow","Green","Blue","Purple","Grey","White"} while True: second_colour = raw_input("\nPlease State The Second Colour Band\nBlack\nBrown\nRed\nOrange\nYellow\nGreen\nBlue\nPurple\nGrey\nWhite\nYour Choice?\n") if second_colour == "Black": second_colour = 0 break elif second_colour == "Brown": second_colour = 1 break elif second_colour == "Red": second_colour = 2 break elif second_colour == "Orange": second_colour = 3 break elif second_colour == "Yellow": second_colour = 4 break elif second_colour == "Green": second_colour = 5 break elif second_colour == "Blue": second_colour = 6 break elif second_colour == "Purple": second_colour = 7 break elif second_colour == "Grey": second_colour = 8 break elif second_colour == "White": second_colour = 9 break else: print "Invalid colour name" third_colours = {"Black","Brown","Red","Orange","Yellow","Green","Blue","Purple","Grey","White"} while True: third_colour = raw_input("\nPlease State The Third Colour Band\nBlack\nBrown\nRed\nOrange\nYellow\nGreen\nBlue\nPurple\nGrey\nWhite\nYour Choice?\n") if third_colour == "Black": third_colour = 1 break elif third_colour == "Brown": third_colour = 10 break elif third_colour == "Red": third_colour = 100 break elif third_colour == "Orange": third_colour = 1000 break elif third_colour == "Yellow": third_colour = 10000 break elif third_colour == "Green": third_colour = 100000 break elif third_colour == "Blue": third_colour = 1000000 break elif third_colour == "Purple": third_colour = 10000000 break elif third_colour == "Grey": third_colour = 100000000 break elif third_colour == "White": third_colour = 1000000000 break else: print "Invalid colour name" fourth_colours = {"Gold","Silver","None"} while True: fourth_colour = raw_input("\nPlease State The Fourth Colour Band\nGold\nSilver\nNone\nYour Choice?\n") if fourth_colour == "Gold": fourth_colour = 5 break elif fourth_colour == "Silver": fourth_colour = 10 break elif fourth_colour == "None": fourth_colour = 20 break else: print "Invalid colour name" total = second_colour*third_colour print "The Total Value Of Your Resistor Is : " + str(first_colour) + str(total) + " +-" + str(fourth_colour) + '%\n' exitprogram = raw_input ("\nPress Enter To Exit")

Yes, solution with nth_element added, so simple. class Solution { public: int findKthLargest(vector& nums, int k) { nth_element(nums.begin(), nums.end() - k, nums.end()); return nums[nums.size() - k]; } };

no problem. Glad it helps

Great... it works!!!! Thanks for your quick help

in conn.php, you have to define your DB constants details such as

Oh! i did not found conn.php file in my PHPBB script.. where it is? mysql_connect('MY DB NAME', 'DB USER NAME', 'DB USER PASS'); mysql_select_db('androidsma_v2'); There i used my real db information

have you put DB details correctly in conn.php? require_once('conn.php'); mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); and, if it is 500 ERR, it usually means you need to check the apache2/nginx server log to find out the exact errors.

HTTP ERROR 500 That error

of course it is working. what errors did you get?

Its not working.. i got error... Did you test it ?

#include #include #include #include const char * colors[10] = { "black", "brown", "red", "orange", "yellow", "green", "blue", "violet", "grey", "white" }; int get_band (int n) { char input[10]; printf ("Enter color of band# %d: " , n+1); scanf ("%s", input); for (int i=0; i<10; i++) { if (strcmp(input, colors[i]) == 0) return i; } printf ("Invalid input\n"); exit (1); } int main() { int band[3]; long value; for (int i=0; i<3; i++) band[i] = get_band(i); value = band[0] * 10 + band[1]; value *= (long)pow(10.0, (double)band[2]); printf ("The value of that resistor is %ld ohms\n", value); system("pause"); return 0; } i don't know how to put a tolerance

what have you done so far?

Project Title: Resistor Calculator Resistor is an electrical component that reduces an electric current. The resistor's ability to reduce current is called resistance and is measured in units of ohms, Ω. The resistance of a resistor and its tolerance are marked on the resistor by color bands that denotes the resistance value as shown in Figure 1. Table 1 shows the value of each colour band. Figure 1: Resistor colour band. Table 1: Digit, multiplier and tolerance band chart. Colour 1st Band 2nd Band 3rd Band (Multiplier) 4th Band (Tolerance) Black 0 0 1 Brown 1 1 10 Red 2 2 100 Orange 3 3 1 000 Yellow 4 4 10 000 Green 5 5 100 000 Blue 6 6 1 000 000 Violet 7 7 10 000 000 Grey 8 8 100 000 000 White 9 9 1 000 000 000 Gold ±5% Silver ±10% None ±20% You are required to design a program to determine the value of a resistor based on resistor colour code. The program should prompt users to request the 4 bands colour marked on the resistor. The result should be display along with its tolerance. The user then asked whether he/she want to continue to determine the value of other resistor or not. If the user enter “Y” or “y”, the program will continue, otherwise enter “N”or “n” the program will stop. The program also should prompt an error message if the user enter an invalid colour code. Bonus marks will be given to any group that can show the correct error message in each selection of colour code. Example calculation By referring to Figure 1, the colour band of the resistor is as follow: 1st band : Red --------> 2 2nd band : Yellow ----> 4 3rd band : Green -----> 100 000 4th band : Silver ------> ±10% Thus, its resistance value is : Formula : [( 10 x 1st band ) + 2nd band ] x 3rd band ) ± 4th band % Resistance = [ (10 x 2 + 4) x 100 000] ±10% = 2 400 000 Ω ±10% please help me to finish this code

yes, you are right.

It can be done in O(N) . Use nth_element() or the divide and conquer approach.

Is this an assignment? 🙂 send it to my address, and I will have a look, but no promise. dr.zhihua.lai [AT] gmail.com

can you help me to do a programing using c++ to determine the colour code of resistor

I see no reason why it cannot

That is interesting. Can users implement destructors in file handling? In other words, can users perform destructive copy in Python?

I now understand your solution, however, according to https://leetcode.com/problems/find-the-duplicate-number/ it has only one duplicate number, but the duplicate number may appear multiple times. int findDuplicate(vector& nums) { int x = 0; for (int i = 0; i < nums.size(); i ++) { x ^= nums[i]; } int y = 0; for (int i = 0; i

I just want to see how narrow of this comment can be.