Codeforces Round #177 (Div. 2)---D. Polo the Penguin and houses (combinatorial Math + violence)

Little Penguin Polo loves his home village. The village has n houses, indexed by integers from 1 to n. Each house had a plaque containing an integer and the I-th house had a plaque containing integer pi (1≤pi≤n).

Little Penguin Polo loves walking around this village. The walk looks like that. First he stands by a house number x. Then he goes to the house whose number was written on the plaque of House X (that's, to house px) and then he goes to the Hou Se whose number is written on the plaque of House px (that's, to house ppx), and so on.

We know that:

When the penguin starts walking is indexed from 1 to K, inclusive, he can walk to house number 1.
When the penguin starts walking from any house indexed from K + 1 to N, inclusive, he definitely cannot walk to house numb ER 1.
When the penguin starts walking from house number 1, he can get back to house number 1 after some non-zero number of walks From a house to a house. 

You need to find the number of ways your may write the numbers on the houses ' plaques so as to fulfill the three above desc ribed conditions. Print the remainder after dividing this number by 1000000007 (109 + 7).
Input

The single line contains the space-separated integers n and k (1≤n≤1000, 1≤k≤min (8, N))-the number of the houses and the number k from the statement.
Output

In a single line print a single integer-the answer to the problem modulo 1000000007 (109 + 7).
Sample Test (s)
Input

5 2

Output

54

Input

7 4

Output

1728

At first I thought the numbers on the labels were different.
Then I read test instructions.
K+1->n and 1, n are separated, the former scheme number is (n−k) (n−k) (n-k) ^ {(n-k)}
Then for 1, there are k schemes, for 2->k, K is very small, 77 7^{7}, direct Search bar

/************************************************************************* > File name:cf-177-d.cpp > Autho R:alex > Mail:zchao1995@gmail.com > Created time:2015 April 08 Wednesday 19:09 18 seconds **************************** /#include <functional> #include <algorithm> #include < iostream> #include <fstream> #include <cstring> #include <cstdio> #include <cmath> #include & lt;cstdlib> #include <queue> #include <stack> #include <map> #include <bitset> #include <

Set> #include <vector> using namespace std;
Const double PI = ACOs (-1.0);
const int inf = 0X3F3F3F3F;
Const double EPS = 1e-15;
typedef long Long LL;

typedef pair &LT;INT, int> PLL;

static const int mod = 1000000007;
LL Res;
int n, K;
int arr[10];

BOOL VIS[10];
        void dfs (int cur) {if (cur > k) {bool flag = 1;
    for (int i = 2; I <= K; ++i) {        memset (Vis, 0, sizeof (VIS));
            Vis[i] = 1;
            int u = arr[i];
                while (U! = 1) {if (Vis[u]) {break;
                } Vis[u] = 1;
            U = arr[u];
                } if (U! = 1) {flag = 0;
            Break
            }} if (flag) {++res;
        Res%= mod;
    } return;
        } for (int i = 1; I <= K; ++i) {arr[cur] = i;
    DFS (cur + 1);
        }} int main () {while (~scanf ("%d%d", &n, &k)) {LL ans = 1;
            for (int i = 1; I <= n-k; ++i) {ans *= (n-k);
        Ans%= MoD;
        } ans *= k;
        Ans%= MoD;
        res = 0;
        DFS (2);
        Ans *= Res;
        Ans%= MoD;
    printf ("%lld\n", ans);
} return 0; }