fzu2282 Wand 数学

Problem 2282 Wand

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）

错位排序

预处理逆元

import java.io.BufferedInputStream;import java.io.PrintWriter;import java.util.Scanner;public class Main {public static void main(String[] args) {new Task().solve();}}class Task {Scanner in = new Scanner(new BufferedInputStream(System.in)) ;PrintWriter out = new PrintWriter(System.out);final long Mod = 1000000007L ;final int N = 10000 ;long[] D = new long[N+1] ;long[] inv = new long[N+1] ;void solve(){D[0] = 1 ;D[1] = 0 ;D[2] = 1 ;for(int i = 3 ; i <= N ; i++){    D[i] = (i-1) * ((D[i-2] + D[i-1]) % Mod) % Mod ;}inv[1] = 1;      for(int i = 2 ; i <= N ; i++){          inv[i] = (Mod - Mod / i) % Mod * inv[(int)Mod % i] % Mod  ;      }  int t = in.nextInt() ;while(t-- > 0){long n = in.nextLong() ;long k = in.nextLong() ;long C = 1 ;for(int i = 1 ; i < k ; i++){C = C * inv[i] % Mod * (n-i+1) % Mod ;}long sum = 0 ; for(int i = (int)k ; i <= n ; i++){C = C * inv[i] % Mod * (n-i+1) % Mod ;sum = sum + D[(int) (n-i)] * C % Mod ;    sum %= Mod ;}out.println(sum) ;}out.flush() ;}}