Adding some more judges, here and there.
[and.git] / UVa / 11282 - Mixing invitations / 11282.cpp
blob5f6549a84c44cc90bca1ef619574ad1e5f192a2c
1 /*
2 Problem: 11282 - Mixing invitations
3 Author: Andrés Mejía-Posada
4 (http://blogaritmo.factorcomun.org)
6 Algoritm: Combinatorics formula
7 */
9 using namespace std;
10 #include <algorithm>
11 #include <iostream>
12 #include <iterator>
13 #include <sstream>
14 #include <fstream>
15 #include <cassert>
16 #include <climits>
17 #include <cstdlib>
18 #include <cstring>
19 #include <string>
20 #include <cstdio>
21 #include <vector>
22 #include <cmath>
23 #include <queue>
24 #include <deque>
25 #include <stack>
26 #include <map>
27 #include <set>
29 const int N = 20;
31 long long choose[N+1][N+1], d[N+1]; //d[i] = Number of derangements if i elements.
33 int main(){
34 /* Binomial coefficients */
35 for (int i=0; i<=N; ++i) choose[i][0] = choose[i][i] = 1;
36 for (int i=1; i<=N; ++i)
37 for (int j=1; j<i; ++j)
38 choose[i][j] = choose[i-1][j-1] + choose[i-1][j];
40 /* Derangements */
41 d[0] = 1, d[1] = 0;
42 for (int i=2; i<=N; ++i) d[i] = (i-1)*(d[i-2] + d[i-1]);
44 int n, m;
45 while (cin >> n >> m){
46 long long ans = 0;
47 for (int k=0; k<=m; ++k) ans += choose[n][k] * d[n-k];
48 cout << ans << endl;
50 return 0;
54 Explanation: Let f(n, k) = Number of permutations of n elements
55 such that exactly k of them are in their correct position. For
56 example, f(3, 1) = 3 which are {1,3,2}, {3,2,1} and {2,1,3}.
57 Then the answer is sum for every 0 <= k <= m of f(n, k).
59 Let D(n) = Numbers of derangenents of n elements. Note that for
60 every n we have f(n, 0) = D(n). And in general,
62 f(n, k) = choose(n, k) * D(n - k)
64 The idea behind this is that we "nail" k elements from the whole
65 sequence of n elements in choose(n, k) ways and for each possi-
66 bility we disorder completely the missing n - k elements in
67 D(n - k) ways.
69 D(i) is computed using a recurrent function, although could also
70 be found using inclusion-exclusion principle.