foj Problem 2282 Wand

 Problem 2282 Wand

Accept: 432    Submit: 1537
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

2
1 1
3 1

Sample Output

1
4

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）

 

题意：有n个法师n根魔法棒,每个法师都有属于自己的1根魔法棒，求至少有k个法师没拿到自己的魔法棒的情况数。

思路：首先题目中k远远小于n，所以可以考虑相反面，考虑只有1,2,...,k-1个法师没拿到自己的魔法棒的情况数，再用n!减去这些情况的情况数即可。

若挑出x个人，这x个人都没拿到自己的魔法棒的情况数记为D(x),即为数量为x个人的错排数，C(n,x)*D(x)就是x个人没拿到魔法棒的总情况数。

AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include <iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#include<bitset>
#include<set>
#include<map>
#include<cmath>
#include<queue>
using namespace std;
#define N_MAX 10000+4
#define MOD 1000000007
#define INF 0x3f3f3f3f
typedef long long ll;
int n, k;
ll dp[N_MAX];//错排数
ll C[N_MAX][ + ];
void init() {
    dp[] = ; dp[] = ; dp[] = ;
    for (int i = ; i < N_MAX; i++) {
        dp[i] = ((i - )*(dp[i - ] + dp[i - ]) + MOD) % MOD;
    }
}
void C_table() {
    for (int i = ; i < N_MAX; i++) {
        C[i][] = ;
        for (int j = ; j <= min(, i); j++) {//!!!j<=i并且数组中j不能超过100
            C[i][j] = (C[i - ][j] + C[i - ][j - ]) % MOD;
        }
    }
}

int main() {
    init(); C_table();
    int t; scanf("%d", &t);
    while (t--) {
        scanf("%d%d", &n, &k);
        ll num = , ans = ;
        for (ll i = ; i <= n; i++) {
            ans = ans*i%MOD;
        }
        for (int i = ; i <= k - ; i++) {
            num = (num + C[n][i] * dp[n - i]) % MOD;
        }
        printf("%lld\n", (ans - num + MOD) % MOD);
    }
    return ;
}