青原樱（组合数学）

Posted 2022-09-30 oneman233

n个位置m盆花，两盆花之间必须有一个空位，问有多少种合法情况。

先考虑把m盆花都放进一个位置里，那么剩下的位置有n-m个，m盆花要插到剩下的位置的空隙当中去，方案数为：

c[n-m+1][m]

又因为每盆花各不相同，最后答案还得乘个m的全排列

化个简发现答案是：

(n-m+1)!/(n-2m+1)!

于是直接一个for循环搞定，如果答案为0的花在for循环中ans一定会被更新为0

代码：

#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=500005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=100000007;
const db eps=1e-10;
int qpow(int a,int b)int tmp=a%MOD,ans=1;while(b)if(b&1)ans*=tmp,ans%=MOD;tmp*=tmp,tmp%=MOD,b>>=1;return ans;
int lowbit(int x)return x&-x;
int max(int a,int b)return a>b?a:b;
int min(int a,int b)return a<b?a:b;
int mmax(int a,int b,int c)return max(a,max(b,c));
int mmin(int a,int b,int c)return min(a,min(b,c));
void mod(int &a)a+=MOD;a%=MOD;
bool chk(int now)
int half(int l,int r)while(l<=r)int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;return l;
int ll(int p)return p<<1;
int rr(int p)return p<<1|1;
int mm(int l,int r)return (l+r)/2;
int lg(int x)if(x==0) return 1;return (int)log2(x)+1;
bool smleql(db a,db b)if(a<b||fabs(a-b)<=eps)return true;return false;
db len(db a,db b,db c,db d)return sqrt((a-c)*(a-c)+(b-d)*(b-d));
bool isp(int x)if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;

int type,n,m,p;

signed main()
    cin>>type>>n>>m>>p;
    int ans=1;
    re(i,m,2*m-1)
        ans*=(n-i+1);
        ans%=p;
    
    cout<<ans;
    return 0;