FZU 2282 Wand 【组合数学】

最新推荐文章于 2019-04-13 15:55:23 发布

Puya

最新推荐文章于 2019-04-13 15:55:23 发布

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分类专栏： HUST 组合数学

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本文探讨了一种计数错位排列的问题，即求解在重新排列一系列魔杖后，至少有k个巫师能拿到自己原来的魔杖的方法数量。通过对全排列进行计算并减去特定条件下不符合要求的排列数，利用动态规划和组合数学技巧得出解决方案。

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Problem 2282 Wand

Accept: 47 Submit: 191
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）

思路：全部的排列减去不符合情况的排列数量（不符合的情况是：0~k-1个东西不在自己的位置上）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
#define ll long long
#define ms(a,b)  memset(a,b,sizeof(a))
#define maxn 510
const int M=1e6+10;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
int i,j,k,n,m;
int r1,r2,l1,l2;
ll d[M];
ll c[M];
ll a[M];
void init()
{
    d[0]=0;d[1]=0;d[2]=1;
    for(int i=3;i<=10000;i++){
        d[i]=(((i-1)%mod)*((d[i-1]+d[i-2])%mod))%mod;
    }
    a[0]=1;
    for(int i=1;i<=10000;i++){
        a[i]=(a[i-1]*i)%mod;
    }
}

ll ex_gcd(ll a,ll b,ll &x,ll &y)
{
    ll d=a;
    if(b!=0){
        d=ex_gcd(b,a%b,y,x);
        y-=(a/b)*x;
    }
    else {
        x=1;
        y=0;
    }
    return d;
}

ll mod_inverse(int b,int mod)
{
    ll x,y;
    ex_gcd(b,mod,x,y);
    return (mod+x%mod)%mod;
}
void solve(int n,int k)
{
    c[0]=1;
    for(int i=1;i<=k;i++){
        ll kk=mod_inverse(i,mod);
        c[i]=(c[i-1]*kk)%mod;
        c[i]=(c[i]*(n-i+1))%mod;
    }
}

int main()
{
    int t;
    scanf("%d",&t);
    init();
    while(t--){
        scanf("%d%d",&n,&k);
        solve(n,k);
        ll ans=0;
        for(int i=0;i<=k-1;i++){
            ans+=((d[n-i])*c[i])%mod;
            ans%=mod;
        }
        ans=(a[n]-ans+mod)%mod;
        printf("%lld\n",ans);
    }
    return 0;
}