# Project Euler with Befunge

# Problem 076: Counting summations

**Description:**

It is possible to write five as a sum in exactly six different ways:

```
4 + 1
3 + 2
3 + 1 + 1
2 + 2 + 1
2 + 1 + 1 + 1
1 + 1 + 1 + 1 + 1
```

How many different ways can one hundred be written as a sum of at least two positive integers?

**Solution:**

v

v

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v+1 p:+3\+1g06:$< > :50g\-3v

>"d"30p1>:30g`#v_:50p060p1>:50g-!#^_::::50g\-`#^_: 3v

>1+:1+g.@ ^+1p+3g05+3\p06:+g06g+3-\g05\+<

**Output:**

**Stack:**

*(0)*

**Explanation:**

The big trick is - similar to many other problems - *caching*.

This problem remembered me a little bit of problem-15.
We use a `100x100`

grid to remember pre-calculated sums.

So in cell [3, 6] is the amount of sums which result in `6`

, start with `3`

and have all summands in descending order:

```
3 + 3
3 + 2 + 1
3 + 1 + 1
```

You see `cache[3,6] = 3`

.

Now to find a new value (for example `[4, 7]`

) we just have to look at the our cache:

```
7 = 4 + x
sum(x) = 7-4 = 3
// first_digit_of_x <= first_digit, because of the descending order
sum(x) = sum([n, 3]); n = [1..3]
sum(x) = [3, 3] + [2, 3] + [1, 3]
```

You see it's important not only to remember the amount but also the first (= highest) summand, so we can guarantee the oder of the sums (an this way that we don't count any sums multiple times).

*Note:* `cache[a, a]`

is always `1`

. But the problem rules dictate that when we calculate the final result we must ignore this (`100 = 100`

is not a valid solution)

Oh and this algorithm improves the native approach (enumerating all solutions) from `O(wtf)`

to `O(n^2)`

.
I'm not sure if I would be still alive when my first algorithm finishes :)

**Edit:**

I did a little optimization:

The value of cell `[d, s]`

is now the sum of all previous cells from `[0, s]`

to `[d, s]`

.

This way we don't have to iterate through all the cells from 0 to d every time. We can just look at the biggest cell which contains the sum of all previous.

Interpreter steps: |
296 178 |

Execution time (BefunExec): |
32ms (9.26 MHz) |

Program size: |
104 x 108 |

Solution: |
190569291 |

Solved at: |
2015-08-26 |