hdu1521 排列组合 指数型母函数模板题

排列组合

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4891    Accepted Submission(s): 2122

Problem Description

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。

 

Input

每组输入数据有两行，第一行是二个数n,m(1<=m,n<=10)，表示物品数，第二行有n个数，分别表示这n件物品的数量。

 

Output

对应每组数据输出排列数。(任何运算不会超出2^31的范围)

 

Sample Input

2 2 1 1

 

Sample Output

2

 

Author

xhd

 

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从题目中可以看出来，取m件物品可能取到重复的物品，所以这是一个指数型母函数问题

接下来就是套用母函数模板了

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iterator>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 2*1e2 + 10;
const int mod = 10000;
typedef long long ll;
ll f( ll x ) {
    ll sum = 1;
    for( ll i = 1; i <= x; i ++ ) {
        sum *= i;
    }
    return sum;
}
int main() {
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    ll n, m, num[maxn];
    double a[maxn], b[maxn];
    while( cin >> n >> m ) {
        memset( a, 0, sizeof(a) );
        memset( b, 0, sizeof(b) );
        for( ll i = 0; i < n; i ++ ) {
            cin >> num[i];
        }
        for( ll i = 0; i <= num[0]; i ++ ) {
            a[i] = 1.0/f(i);
        }
        for( ll i = 1; i < n; i ++ ) {
            for( ll j = 0; j <= n; j ++ ) {
                for( ll k = 0; k <= num[i] && k+j <= n; k ++ ) {
                    b[k+j] += a[j]/f(k);
                }
            }
            for( ll j = 0; j <= n; j ++ ) {
                a[j] = b[j], b[j] = 0;
            }
        }
        printf("%.lf
",a[m]*f(m) );
    }
    return 0;
}