izanbf1803

#include <iostream>
#include <iomanip>
#include <vector>
#include <unordered_set>
#include <algorithm>
#include <stdlib.h> // qsort
#include <chrono>
#include <cassert>

using std::chrono::high_resolution_clock;

izanbf1803

from typing import TextIO
from datetime import date
import time
import requests
import json
import os
import sys
import re

izanbf1803

Keybase proof

I hereby claim:

• I am izanbf1803 on github.
• I am izanbf (https://keybase.io/izanbf) on keybase.
• I have a public key ASCMeN20EyeipWshitqPFu-yVZZSPv6gnqJvm3cmo5GKago

To claim this, I am signing this object:

izanbf1803

# Exemple per executar: 
#   python simulate.py Stonks Stonks Stonks2 Stonks2 100

import subprocess
import random
import os
import sys
from multiprocessing import Pool, Lock, current_process, Value
from collections import defaultdict

izanbf1803

# Exemple per executar: 
#   python simulate.py Stonks Stonks Stonks2 Stonks2
# Sortida:
#   Stonks won with seeds: 43913 70491 41637 65191
# 
#   Stonks2 won with seeds: 39428 20517 29384 42524 24253 54463
# 
#   Stonks won 4 times 40.00%
#   Stonks2 won 6 times 60.00%

izanbf1803

#ifndef __VEC_HPP__
#define __VEC_HPP__

#include <iostream>
#include <cmath>

namespace vec
{
    typedef long double ld;
 

izanbf1803

#include <bits/stdc++.h>
using namespace std;

// Khun algorithm, computes maximum bipartite matching in O(|V|^3)

bool tryKhun(int n, const vector<vector<int>>& adj, vector<bool>& used, vector<int>& M, int v)
{
    if (used[v]) return false;
    used[v] = true;
    for (int u : adj[v]) {

izanbf1803

typedef unsigned int uint;

uint popcount_short[0xffff];

void popcount_setup()
{
    popcount_short[0] = 0;
    for (uint mask = 1; mask < 0xffff; ++mask) {
        popcount_short[mask] = (mask&1) + popcount_short[mask>>1];
    }

izanbf1803

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

// This implementation uses cast to __int128 for products because
// of big input (0 < q < 5*10^10).

ll mpow(ll x, ll n, ll m)
{

izanbf1803

import numpy as np
n = int(input())
M = np.matrix([[1,1],[1,0]], dtype="object")
ans = (M**n)[0,0]
print(ans)