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