tpopp

Keybase proof

I hereby claim:

• I am tpopp on github.
• I am tpopp (https://keybase.io/tpopp) on keybase.
• I have a public key whose fingerprint is 79E8 D3A5 952F EBD5 2E2E 2391 88D0 D341 6D3A D349

To claim this, I am signing this object:

tpopp

import java.util.Scanner;
//http://codeforces.com/problemset/problem/371/B


public class Fox {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int[][] facts = new int[2][4];
		facts[0][3] = sc.nextInt();

tpopp

import java.util.Stack;
import java.util.HashMap;
import java.io.*;

public class PotD_4 {

    public static void main(String[] args) throws IOException {
		HashMap<String, Integer> compare;
		compare = new HashMap<>();
		compare.put("+", 1);

tpopp

# improve A(n)
# H(n) improves by working from both sides in at the same time

def A(n):
    res = 0
    for i in range(1,n+1):
       res += n // i
    return res

def H(n):

tpopp

/*
Most comments are obvious; akward structure due to quick coding
Recursion would have made this much easier to follow

Concept:  A,BCD = A,000 + B00 + C0 + D

All numbers will occur in less significant digit based on only the 'letter'
. For every 10, each digit will occur once in the 1's spot;
. For every 100, each digit will occur once in the 1's spot for every 10 (adds up to 10), and will be in tens spot for 10; etc 
All += C * 1 * 10^1; All += B * 2 * 10^2; All += A * 3 * 10^3; etc.

tpopp

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

int count[10]; //count each digit, all initialized to 0

main() {

tpopp

import sys
values = {1:0}

def cycle(i, j):
    large = 0
    for k in range(min(i,j), max(i,j)+1):
        a = compute(k)
        if (a > large):
            large = a
    print(i, j, large+1)  #(large + 1) because I wasn't including 1 in the length of the cycle.