AtCoder - ABC185C. Duodecim Ferra

You can practice the problem Duodecim Ferra [here](https://atcoder.jp/contests/abc185/tasks/abc185_c). ## Problem Statement There is an iron bar of length ``L`` lying east-west. We will cut this bar at 11 positions to divide it into 12 bars. Here, each of the 12 resulting bars must have a positive integer length. Find the number of ways to do this division. Two ways to do the division are considered different if and only if there is a position cut in only one of those ways. Under the constraints of this problem, it can be proved that the answer is less than 2^63. ## Solutions This problem can be solved using Stars and Bars Theorem. Given the lenght ``L``, we have a Diophantine equation ``x[0] + x[1] + ... + x[11] = L`` where x[i] are the lengths of the bars after the division. We need to select 11 positions out of ``L - 1`` positions. For example, if ``L`` is 14, we will have the following possible points to cut. ``` 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 ``` Therefore, the answer is ``` C(n, r) = C(L - 1, r - 1) = C(14 - 1, 12 - 1) = C(13, 11) = 78 ``` We can solve it in O(r) time complexity and O(1) space complexity . nCr can be written as ``` (n)! / (r)!( / (n - r)! = (n * (n - 1) * (n - 2) * ... * 1 ) / (r * (r - 1) * (r - 2) * ... * 1) / ((n - r) * (n - r - 1) * (n - r - 2) * ... * 1) = n * (n - 1) * (n - 2) * ... * (n - (r - 1)) / (r * (r - 1) * (r - 2) * ... * 1) ``` ```cpp // AC - 3 ms void solve() { ll L; cin >> L; ll ans = 1; FOR(i, 1, 12) { ans *= L - i; // n * (n - 1) * (n - 2) * ... * (n - (r - 1)) ans /= i; // r * (r - 1) * (r - 2) * ... * 1 } OUT(ans); } ``` We can turn it to a template for similar problems ```cpp template< typename T > T comb(int64_t N, int64_t K) { if(K < 0 || N < K) return 0; T ret = 1; for(T i = 1; i <= K; ++i) { ret *= N--; ret /= i; } return ret; } // AC - 7 ms void solve() { ll L; cin >> L; OUT(comb<ll>(L - 1, 11)); } ``` We can also use dynamic programming to solve this problem. The recursive formula is ``` C(n, r) = C(n - 1, r - 1) + C(n - 1, r) ``` For ``r == 0`` and ``n == r``, the result would be 1. ```cpp // AC - 10 ms const int mxN = 205; ll c[mxN][mxN]; void solve() { ll L; cin >> L; REP(i, l) { c[i][0] = c[i][i] = 1; FOR(j, 1, i) { c[i][j] = c[i - 1][j - 1] + c[i - 1][j]; } } OUT(c[L - 1][11]); } ```