hdu 5894 hannnnah_j’s Biological Test 组合数学

传送门：hdu 5894 hannnnah_j’s Biological Test

题目大意：n个座位，m个学生，使每个学生的间隔至少为k个座位

组合中的插空法

思路：每个学生先去掉k个空位间隔，剩下n-k*m；这些空位至少要坐m个学生，n-k*m-1个空，插m-1个门，方法数为：c(n-k*m-1,m-1)；当只有一个学生时，间隔K个位的条件就没必要了，也就是n>k+1的条件不一定要成立

顺带弄了个Lucas的模板

/**************************************************************
    Problem:hdu 5894 hannnnah_j’s Biological Test
    User: youmi
    Language: C++
    Result: Accepted
    Time:436MS
    Memory:17240K
****************************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#include<bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <cmath>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#define zeros(a) memset(a,0,sizeof(a))
#define ones(a) memset(a,-1,sizeof(a))
#define sc(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d%d",&a,&b)
#define sc3(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define scs(a) scanf("%s",a)
#define sclld(a) scanf("%I64d",&a)
#define pt(a) printf("%d\n",a)
#define ptlld(a) printf("%I64d\n",a)
#define rep(i,from,to) for(int i=from;i<=to;i++)
#define irep(i,to,from) for(int i=to;i>=from;i--)
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define lson (step<<1)
#define rson (lson+1)
#define eps 1e-6
#define oo 0x3fffffff
#define TEST cout<<"*************************"<<endl
const double pi=4*atan(1.0);

using namespace std;
typedef long long ll;
template <class T> inline void read(T &n)
{
    char c; int flag = 1;
    for (c = getchar(); !(c >= ‘0‘ && c <= ‘9‘ || c == ‘-‘); c = getchar()); if (c == ‘-‘) flag = -1, n = 0; else n = c - ‘0‘;
    for (c = getchar(); c >= ‘0‘ && c <= ‘9‘; c = getchar()) n = n * 10 + c - ‘0‘; n *= flag;
}
ll Pow(ll base, ll n, ll mo)
{
    ll res=1;
    while(n)
    {
        if(n&1)
            res=res*base%mo;
        n>>=1;
        base=base*base%mo;
    }
    return res;
}
//***************************

ll n,m,k;
const int maxn=1000000+10;
const ll mod=1000000007;
ll fac[maxn],inver[maxn];
ll inv(ll aa)
{
    return Pow(aa,mod-2,mod);
}
void init()
{
    fac[0]=1,fac[1]=1;
    inver[0]=1,inver[1]=1;
    rep(i,2,maxn-10)
        fac[i]=fac[i-1]*i%mod,inver[i]=inver[i-1]*inv(i)%mod;
}
ll C(ll aa,ll bb)
{
    if(bb>aa)
        return 0;
    ll temp=fac[aa]*inver[bb]%mod*inver[aa-bb]%mod;
    return temp;
}
ll lucas(ll aa,ll bb)
{
    if(bb==0)
        return 1;
    return   C(aa%mod,bb%mod)*lucas(aa/mod,bb/mod)%mod;
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    int T_T;
    scanf("%d",&T_T);
    init();
    for(int kase=1;kase<=T_T;kase++)
    {
        sclld(n),sclld(m),sclld(k);
        ll ans=lucas(n-k*m-1,m-1)*n%mod*inv(m)%mod;
        ptlld(ans);
    }
}