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

# Problem 31: Coin sums

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

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).

# Interactive test

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

Number of test cases (1-5):

Input data (separated by spaces or newlines):
Note: please enter in pence, not pound

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)

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.

       #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.

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++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

# Hackerrank

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.

# Heatmap

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

 green solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too yellow solutions score less than 100% at Hackerrank (but still solve the original problem easily) gray problems are already solved but I haven't published my solution yet blue solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much orange problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte red problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too black problems are solved but access to the solution is blocked for a few days until the next problem is published [new] the flashing problem is the one I solved most recently

I stopped working on Project Euler problems around the time they released 617.
 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
 501 502 503 504 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 520 521 522 523 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 540 541 542 543 544 545 546 547 548 549 550 551 552 553 554 555 556 557 558 559 560 561 562 563 564 565 566 567 568 569 570 571 572 573 574 575 576 577 578 579 580 581 582 583 584 585 586 587 588 589 590 591 592 593 594 595 596 597 598 599 600
 601 602 603 604 605 606 607 608 609 610 611 612 613 614 615 616 617 618 619 620 621 622 623 624 625 626 627 628 629 630 631 632 633 634 635 636 637 638 639 640 641 642 643 644 645 646 647 648 649 650 651 652 653 654 655 656 657 658 659 660 661 662 663 664 665 666 667 668 669 670 671 672 673 674 675 676 677 678 679 680 681 682 683 684 685 686 687 688 689 690 691 692 693 694 695 696 697 698 699 700 701 702 703 704 705 706 707 708
The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

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

 << problem 30 - Digit fifth powers Pandigital products - problem 32 >>
more about me can be found on my homepage, especially in my coding blog.
some names mentioned on this site may be trademarks of their respective owners.
thanks to the KaTeX team for their great typesetting library !