排列组合 HDU-1521 母函数

HDU-1521

有 n 种物品，并且知道每种物品的数量。要求从中选出 m 件物品的排列数。例如有两种物品A,B，并且数量都是1，从中选2件物品，则排列有"AB","BA"两种。

$1\le n,m\le 10$

input

2 2
1 1

output

2

solution

$指数型母函数，\prod _{i=1}^n\sum _{j=0}^{a_1}\frac{x^j}{j!},x^m的系数为答案$

code

//Siberian Squirrel
//#include
#include
#include
#include
#include
#include

using namespace std;
typedef long long ll;

const double PI = acos(-1);
const double eps = 1e-7;
const int MOD = 3221225473;
const int N = 5e6 + 10;

int limit;
ll f[20], a[20];
ll p[20];

inline ll solve(int n, int m, ll res = 0) {
     
    memset(f, 0, sizeof f);
    f[0] = 1;
    for(int i = 1; i <= n; ++ i) {
     
        scanf("%d", &limit);
        memset(a, 0, sizeof a);
        for(int i = m; i >= 0; -- i) {
     
            for (int j = 1; i + j <= m && j <= limit; ++ j) {
     
                a[j + i] += p[j + i] * f[i] / p[j] / p[i];
            }
        }
        for(int j = 1; j <= m; ++ j)
            f[j] += a[j];
    }

    return f[m];
}

int main() {
     
#ifdef ACM_LOCAL
    freopen("input", "r", stdin);
    freopen("output", "w", stdout);
#endif
    int o = 1, n, m;
//    scanf("%d", &o);
    p[0] = p[1] = 1;
    for(int i = 2; i <= 12; ++ i) p[i] = p[i - 1] * i;
    while(o --) {
     
        while(~scanf("%d%d", &n, &m))
            printf("%lld\n", solve(n, m));
    }
    return 0;
}