FZU 2282 Wand (组合数学 错排应用)

 Problem 2282 Wand

Accept: 83    Submit: 305
Time Limit: 1000 mSec    Memory Limit : 262144 KB

 Problem Description

N wizards are attending a meeting. Everyone has his own magic wand. N magic wands was put in a line, numbered from 1 to n(Wand_i owned by wizard_i). After the meeting, n wizards will take a wand one by one in the order of 1 to n. A boring wizard decided to reorder the wands. He is wondering how many ways to reorder the wands so that at least k wizards can get his own wand.

For example, n=3. Initially, the wands are w1 w2 w3. After reordering, the wands become w2 w1 w3. So, wizard 1 will take w2, wizard 2 will take w1, wizard 3 will take w3, only wizard 3 get his own wand.

 Input

First line contains an integer T (1 ≤ T ≤ 10), represents there are T test cases.

For each test case: Two number n and k.

1<=n <=10000.1<=k<=100. k<=n.

 Output

For each test case, output the answer mod 1000000007(10^9 + 7).

 Sample Input

21 13 1

 Sample Output

14

 Source

第八届福建省大学生程序设计竞赛-重现赛（感谢承办方厦门理工学院）

题目链接：http://acm.fzu.edu.cn/problem.php?pid=2282

题目分析：题目意思就是n个数字里要求至少k个数字位置不变，其余进行错排的方案数，容易想到不变的数字从k枚举到n，每次取i(k <= i < n)个出来，对剩下的n-i个进行错排，即C(n, i) * dp[n - i] (dp[n - i]表示对n-i个数进行错排的方案数)，然后将它们累加，但是这样做超时了。

考虑到n较大，k较小，可以反过来处理，总的方案数为n!，可以从总的方案数里减去不符合条件的情况，即不变的个数从0枚举到k-1

#include <cstdio>#include <cstring>#include <algorithm>#define ll long longusing namespace std;int const MOD = 1e9 + 7;int n, K;ll dp[10005], fac[10005];ll qpow(ll a, ll x) {      ll res = 1;      while(x) {          if (x & 1) {              res = (res * a) % MOD;          }          a = (a * a) % MOD;          x >>= 1;      }      return res;  }    ll C(ll n, ll k) {        if(n < k) {          return 0;      }      if(k > n - k) {            k = n - k;        }      ll a = 1, b = 1;        for(int i = 0; i < k; i++) {            a = a * (n - i) % MOD;          b = b * (i + 1) % MOD;      }        return a * qpow(b, MOD - 2) % MOD;  }  void pre() {dp[0] = 1;dp[1] = 0;dp[2] = 1;fac[1] = 1;fac[2] = 2;for (int i = 3; i <= 10000; i ++) {dp[i] = (((i - 1) % MOD) * ((dp[i - 2] + dp[i - 1])) % MOD) % MOD;fac[i] = ((fac[i - 1] % MOD) * (i % MOD)) % MOD;}}  int main() {pre();int T;scanf("%d", &T);while (T --) {scanf("%d %d", &n, &K);ll ans = 0;for (int i = 0; i < K; i ++) {ans = ((ans % MOD) + (C(n, i) * dp[n - i]) % MOD) % MOD;}printf("%I64d\n", (MOD + (fac[n] % MOD) - (ans % MOD)) % MOD);}}