FZU 2282 Wand 【组合数学】

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 510const 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;}