BZOJ 2467 生成树(组合数学)

题意：求n-五边形的生成树个数。

结论题，答案为4*n*5^(n-1).

首先中心的n边形一定需要切掉一个边，C(1,n).

然后每个五边形都切一个边，C(1,4)*5^(n-1).

于是答案就是4*n*5^(n-1).

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <bitset>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-8
# define MOD 2007
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
const int N=105;
//Code begin...

int mod[N];

int main ()
{
    int T, n;
    scanf("%d",&T);
    mod[0]=1; FOR(i,1,100) mod[i]=mod[i-1]*5%MOD;
    while (T--) {
        scanf("%d",&n);
        printf("%d\n",4*n*mod[n-1]%MOD);
    }
    return 0;
}