Solução por Lucca Siaudzionis
Vamos definir por $$f(b, g)$$ o número de maneiras formar os pares com $$b$$ meninos e $$g$$ meninas. Primeiro, vamos notar que $$f(x, x) = x! $$. Suponhamos agora $$b > g$$. Vamos, fixado o número de garotas, achar calcular todos os $$f(b, g)$$:
$$!f(b, g) = g \times ( \ f(b-1, g) + f(b-1, g-1)\ )$$
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| // | |
| // baile_de_formatura.cpp | |
| // | |
| // Created by Lucca Siaudzionis on 19/11/14. | |
| // Copyright (c) 2014 Luccasiau. All rights reserved. | |
| // | |
| #include <cstdio> | |
| #define MOD 1000000007 | |
| typedef long long lli; | |
| #define MAXN 1010 | |
| int g, b; | |
| lli fact[MAXN]; | |
| lli ways[MAXN][MAXN]; | |
| int main(){ | |
| fact[0] = 1LL; | |
| for(int i = 1;i < MAXN;i++) | |
| fact[i] = (lli(i)*fact[i-1]) % MOD; | |
| for(int i = 1;i < MAXN;i++) | |
| ways[1][i] = 1LL, | |
| ways[i][i] = fact[i]; | |
| for(int i = 2;i < MAXN;i++) | |
| for(int j = i+1;j < MAXN;j++) | |
| ways[i][j] = (lli(i)*( ways[i][j-1] + ways[i-1][j-1] )) % MOD; | |
| while(scanf("%d %d", &b, &g) && b) printf("%lld\n", ways[g][b]); | |
| return 0; | |
| } |
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