<< problem 30 - Digit fifth powers | Pandigital products - problem 32 >> |

# Problem 31: Coin sums

(see projecteuler.net/problem=31)

In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation:

1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).

It is possible to make £2 in the following way:

1x £1 + 1x 50p + 2x 20p + 1x 5p + 1x 2p + 3x 1p

How many different ways can £2 be made using any number of coins?

# My Algorithm

My program creates a table `history`

that contains the number of combinations for a given sum of money:

- its entry `history[0] `

refers to £0

- its entry `history[1]`

refers to £0.01

- its entry `history[2]`

refers to £0.02

- its entry `history[3]`

refers to £0.03

- ...

- its entry `history[200]`

refers to £2.00

There are 8 different coins and therefore each entry of `history`

is a `std::vector`

itself with 8 elements:

it tells how many combinations exist if only the *current coin or smaller coins* are used.

For example, there is always one way/combination to pay a certain amout if you only have single pennies.

That means, the first element is always `1`

.

Moreover, each of the next element is at least 1, too, because I said: "current coin *or smaller coins*".

If we would like to pay £0.01 then `history[1] = { 1,1,1,1,1,1,1,1 }`

.

Now comes the only part that isn't obvious: there is one combination of paying *zero pounds*, too:

`history[0] = { 1,1,1,1,1,1,1,1 }`

. From now on, everything comes natural, trust me, ...

If we would like to pay £0.02 then there are two ways: pay with two single pennies or a 2p coin.

What we do is:

1. try *not* to use the current coin (2p in our case), only smaller coins → there is one combination

2. try to use the current coin (2p in our case) → then there are 0.00 £ left which is possible in one way

3. add 1.+2.

So far we had `history[2] = { 1,?,?,?,?,?,?,? }`

Step 1 is the same as `history[2][currentCoinId - 1] = history[2][0] = 1`

.

Step 2 is the same as `history[2 - currentCoinValue][currentCoinId] = history[0][1] = 1`

.

Therefore we have `1+1=2`

combinations (as expected:) `history[2] = { 1,2,?,?,?,?,?,? }`

.

The next coin, it's the 5p coin, can't be used because it's bigger than the total of £0.02. In software terms `currentCoinValue > total`

.

Only step 1 applies to all remaining elements: `history[2] = { 1,2,2,2,2,2,2,2 }`

.

What does it mean ? There are 2 ways to pay 0.02 £ with 1p and 2p. And there are still only two ways if you use all coins up to £2.

When the program computes `history[200]`

then the result of the problem is stored in the last element (`history[200][7]`

).

## Modifications by HackerRank

There are multiple test cases. My program computes all combinations up to the input values and stores them in `history`

.

If a test case's input is smaller than something we had before then no computation is required at all, it will become a basic table lookup.

The results may exceed 32 bits and thus I compute mod 10^9+7 whenever possible (as requested by their modified problem statement).

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

#include <iostream>
#include <vector>
const unsigned int NumCoins = 8;
// face value of all coins in cents

const unsigned int Coins[NumCoins] = { 1,2,5,10,20,50,100,200 };
// store number of combinations in [x] if coin[x] is allowed:
// [0] => combinations if only pennies are allowed
// [1] => 1 cent and 2 cents are allowed, nothing more
// [2] => 1 cent, 2 cents and 5 cents are allowed, nothing more
// ...
// [6] => all but 2 pounds (= 200 cents) are allowed
// [7] => using all coins if possible

typedef std::vector<unsigned long long> Combinations;
int main()
{
// remember combinations for all prices from 1 cent up to 200 cents (2 pounds)
std::vector<Combinations> history;
unsigned int tests;
std::cin >> tests;
while (tests--)
{
unsigned int total;
std::cin >> total;
// initially we start at zero
// but if there are previous test cases then we can re-use the old results
for (unsigned int cents = history.size(); cents <= total; cents++)
{
// count all combinations of those 8 coins
Combinations ways(NumCoins);
// one combination if using only 1p coins (single pennys)
ways[0] = 1;
// use larger coins, too
for (size_t i = 1; i < ways.size(); i++)
{
// first, pretend not to use that coin (only smaller coins)
ways[i] = ways[i - 1];
// now use that coin once (if possible)
auto currentCoin = Coins[i];
if (cents >= currentCoin)
{
auto remaining = cents - currentCoin;
ways[i] += history[remaining][i];
}
// not needed for the original problem, only for Hackerrank's modified problem
ways[i] %= 1000000007;
}
// store information for future use
history.push_back(ways);
}
// look up combinations
auto result = history[total];
// the last column (allow all coins) contains the desired value
auto combinations = result.back();
combinations %= 1000000007; // for Hackerrank only
std::cout << combinations << std::endl;
}
return 0;
}

This solution contains 12 empty lines, 20 comments and 2 preprocessor commands.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

This is equivalent to`echo "1 10" | ./31`

Output:

*Note:* the original problem's input `200`

__cannot__ be entered

because just copying results is a soft skill reserved for idiots.

*(this interactive test is still under development, computations will be aborted after one second)*

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.

(compiled for x86_64 / Linux, GCC flags: `-O3 -march=native -fno-exceptions -fno-rtti -std=c++11 -DORIGINAL`

)

See here for a comparison of all solutions.

**Note:** interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without `-DORIGINAL`

.

# Changelog

February 23, 2017 submitted solution

April 6, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler031

My code solves **9** out of **9** test cases (score: **100%**)

# Difficulty

Project Euler ranks this problem at **5%** (out of 100%).

Hackerrank describes this problem as **easy**.

*Note:*

Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.

In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.

# Links

projecteuler.net/thread=31 - **the** best forum on the subject (*note:* you have to submit the correct solution first)

Code in various languages:

Python: www.mathblog.dk/project-euler-31-combinations-english-currency-denominations/ (written by Kristian Edlund)

Haskell: github.com/nayuki/Project-Euler-solutions/blob/master/haskell/p031.hs (written by Nayuki)

Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p031.java (written by Nayuki)

Mathematica: github.com/nayuki/Project-Euler-solutions/blob/master/mathematica/p031.mathematica (written by Nayuki)

C: github.com/eagletmt/project-euler-c/blob/master/30-39/problem31.c (written by eagletmt)

Go: github.com/frrad/project-euler/blob/master/golang/Problem031.go (written by Frederick Robinson)

Scala: github.com/samskivert/euler-scala/blob/master/Euler031.scala (written by Michael Bayne)

# Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.

yellow problems score less than 100% at Hackerrank (but still solve the original problem).

gray problems are already solved but I haven't published my solution yet.

blue problems are solved and there wasn't a Hackerrank version of it at the time I solved it or I didn't care about it because it differed too much.

red problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte.

*Please click on a problem's number to open my solution to that problem:*

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 |

26 | 27 | 28 | 29 | 30 | 31 | 32 | 33 | 34 | 35 | 36 | 37 | 38 | 39 | 40 | 41 | 42 | 43 | 44 | 45 | 46 | 47 | 48 | 49 | 50 |

51 | 52 | 53 | 54 | 55 | 56 | 57 | 58 | 59 | 60 | 61 | 62 | 63 | 64 | 65 | 66 | 67 | 68 | 69 | 70 | 71 | 72 | 73 | 74 | 75 |

76 | 77 | 78 | 79 | 80 | 81 | 82 | 83 | 84 | 85 | 86 | 87 | 88 | 89 | 90 | 91 | 92 | 93 | 94 | 95 | 96 | 97 | 98 | 99 | 100 |

101 | 102 | 103 | 104 | 105 | 106 | 107 | 108 | 109 | 110 | 111 | 112 | 113 | 114 | 115 | 116 | 117 | 118 | 119 | 120 | 121 | 122 | 123 | 124 | 125 |

126 | 127 | 128 | 129 | 130 | 131 | 132 | 133 | 134 | 135 | 136 | 137 | 138 | 139 | 140 | 141 | 142 | 143 | 144 | 145 | 146 | 147 | 148 | 149 | 150 |

151 | 152 | 153 | 154 | 155 | 156 | 157 | 158 | 159 | 160 | 161 | 162 | 163 | 164 | 165 | 166 | 167 | 168 | 169 | 170 | 171 | 172 | 173 | 174 | 175 |

176 | 177 | 178 | 179 | 180 | 181 | 182 | 183 | 184 | 185 | 186 | 187 | 188 | 189 | 190 | 191 | 192 | 193 | 194 | 195 | 196 | 197 | 198 | 199 | 200 |

201 | 202 | 203 | 204 | 205 | 206 | 207 | 208 | 209 | 210 | 211 | 212 | 213 | 214 | 215 | 216 | 217 | 218 | 219 | 220 | 221 | 222 | 223 | 224 | 225 |

226 | 227 | 228 | 229 | 230 | 231 | 232 | 233 | 234 | 235 | 236 | 237 | 238 | 239 | 240 | 241 | 242 | 243 | 244 | 245 | 246 | 247 | 248 | 249 | 250 |

251 | 252 | 253 | 254 | 255 | 256 | 257 | 258 | 259 | 260 | 261 | 262 | 263 | 264 | 265 | 266 | 267 | 268 | 269 | 270 | 271 | 272 | 273 | 274 | 275 |

276 | 277 | 278 | 279 | 280 | 281 | 282 | 283 | 284 | 285 | 286 | 287 | 288 | 289 | 290 | 291 | 292 | 293 | 294 | 295 | 296 | 297 | 298 | 299 | 300 |

301 | 302 | 303 | 304 | 305 | 306 | 307 | 308 | 309 | 310 | 311 | 312 | 313 | 314 | 315 | 316 | 317 | 318 | 319 | 320 | 321 | 322 | 323 | 324 | 325 |

326 | 327 | 328 | 329 | 330 | 331 | 332 | 333 | 334 | 335 | 336 | 337 | 338 | 339 | 340 | 341 | 342 | 343 | 344 | 345 | 346 | 347 | 348 | 349 | 350 |

351 | 352 | 353 | 354 | 355 | 356 | 357 | 358 | 359 | 360 | 361 | 362 | 363 | 364 | 365 | 366 | 367 | 368 | 369 | 370 | 371 | 372 | 373 | 374 | 375 |

376 | 377 | 378 | 379 | 380 | 381 | 382 | 383 | 384 | 385 | 386 | 387 | 388 | 389 | 390 | 391 | 392 | 393 | 394 | 395 | 396 | 397 | 398 | 399 | 400 |

401 | 402 | 403 | 404 | 405 | 406 | 407 | 408 | 409 | 410 | 411 | 412 | 413 | 414 | 415 | 416 | 417 | 418 | 419 | 420 | 421 | 422 | 423 | 424 | 425 |

426 | 427 | 428 | 429 | 430 | 431 | 432 | 433 | 434 | 435 | 436 | 437 | 438 | 439 | 440 | 441 | 442 | 443 | 444 | 445 | 446 | 447 | 448 | 449 | 450 |

451 | 452 | 453 | 454 | 455 | 456 | 457 | 458 | 459 | 460 | 461 | 462 | 463 | 464 | 465 | 466 | 467 | 468 | 469 | 470 | 471 | 472 | 473 | 474 | 475 |

476 | 477 | 478 | 479 | 480 | 481 | 482 | 483 | 484 | 485 | 486 | 487 | 488 | 489 | 490 | 491 | 492 | 493 | 494 | 495 | 496 | 497 | 498 | 499 | 500 |

501 | 502 | 503 | 504 | 505 | 506 | 507 | 508 | 509 | 510 | 511 | 512 | 513 | 514 | 515 | 516 | 517 | 518 | 519 | 520 | 521 | 522 | 523 | 524 | 525 |

526 | 527 | 528 | 529 | 530 | 531 | 532 | 533 | 534 | 535 | 536 | 537 | 538 | 539 | 540 | 541 | 542 | 543 | 544 | 545 | 546 | 547 | 548 | 549 | 550 |

I scored 13,183 points (out of 15300 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

Look at my progress and performance pages to get more details.

My username at Project Euler is

**stephanbrumme**while it's stbrumme at Hackerrank.

# Copyright

I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.

All of my solutions can be used for any purpose and I am in no way liable for any damages caused.

You can even remove my name and claim it's yours. But then you shall burn in hell.

The problems and most of the problems' images were created by Project Euler.

Thanks for all their endless effort.

<< problem 30 - Digit fifth powers | Pandigital products - problem 32 >> |