{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Euler Problem 31\n",
    "\n",
    "In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation:\n",
    "\n",
    "1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).\n",
    "\n",
    "It is possible to make £2 in the following way:\n",
    "\n",
    "1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p\n",
    "\n",
    "How many different ways can £2 be made using any number of coins?"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "Let's do it non-recursive. Tricky part is to use ceil, otherwise we get wrong ranges for example for $\\frac{50}{20}$."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[0, 1]\n",
      "[0, 1, 2]\n"
     ]
    }
   ],
   "source": [
    "from math import ceil\n",
    "\n",
    "print(list(range(50 // 20)))\n",
    "print(list(range(ceil(50 / 20))))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "from math import ceil\n",
    "\n",
    "r = 200\n",
    "c = 0\n",
    "\n",
    "if r % 200 == 0:\n",
    "    c += 1\n",
    "for p200 in range(ceil(r / 200)):\n",
    "    r200 = r - p200 * 200\n",
    "    if r200 % 100 == 0:\n",
    "        c += 1\n",
    "    for p100 in range(ceil(r200 / 100)):\n",
    "        r100 = r200 - p100 * 100\n",
    "        if r100 % 50 == 0:\n",
    "            c += 1\n",
    "        for p50 in range(ceil(r100 / 50)):\n",
    "            r50 = r100 - p50 * 50\n",
    "            if r50 % 20 == 0:\n",
    "                c += 1\n",
    "            for p20 in range(ceil(r50 / 20)):\n",
    "                r20 = r50 - p20 * 20\n",
    "                if r20 <= 0:\n",
    "                    break\n",
    "                if r20 % 10 == 0:\n",
    "                    c += 1\n",
    "                for p10 in range(ceil(r20 / 10)):\n",
    "                    r10 = r20 - p10 * 10\n",
    "                    if r10 % 5 == 0:\n",
    "                        c += 1\n",
    "                    for p5 in range(ceil(r10 / 5)):\n",
    "                        r5 = r10 - p5 * 5\n",
    "                        if r5 % 2 == 0:\n",
    "                            c += 1\n",
    "                        for p2 in range(ceil(r5 / 2)):\n",
    "                            r2 = r5 - p2 * 2\n",
    "                            if r2 % 1 == 0:\n",
    "                                c += 1\n",
    "\n",
    "s = c"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "73682\n"
     ]
    }
   ],
   "source": [
    "print(s)\n",
    "assert(s == 73682)"
   ]
  }
 ],
 "metadata": {
  "completion_date": "Fri, 25 Aug 2017, 13:02",
  "kernelspec": {
   "display_name": "Python 3",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.5.4"
  },
  "tags": [
   "recursion",
   "coins"
  ]
 },
 "nbformat": 4,
 "nbformat_minor": 2
}