How Many Sets II

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Time Limit: 2 Seconds      

Memory Limit: 65536 KB

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Given a set S = {1, 2, ..., n}, number m and p, your job is to count how many set T satisfies the following condition:

• T is a subset of S
• |T| = m
• T does not contain continuous numbers, that is to say x and x+1 can not both in T

 

Input

There are multiple cases, each contains 3 integers n ( 1 <= n <= 10⁹ ), m ( 0 <= m <= 10⁴, m <= n ) and p ( p is prime, 1 <= p <= 10⁹ ) in one line seperated by a single space, proceed to the end of file.

Output

Output the total number mod p.

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Lucas定理p为质数情况裸题

因为是选的元素不能连续，我们先把选的元素拿出来，剩下的元素有n-m+1个空，选m个插进去行了

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std;typedef long long ll;ll n,m,P;ll Pow(ll a,ll b){    ll ans=1;    for(;b;b>>=1,a=a*a%P)        if(b&1) ans=ans*a%P;    return ans;}ll Inv(ll a){
    return Pow(a,P-2);}ll C(ll n,ll m){    if(n
         
        
       
      
     

 BZOJ 2982: combination

模数10007很小，可以直接线性预处理阶乘和逆元，48ms-->4ms

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std;typedef long long ll;const int N=10007;inline ll read(){    char c=getchar();ll x=0,f=1;    while(c<'0'||c>'9'){
   if(c=='-')f=-1;c=getchar();}    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}    return x*f;}ll n,m,P=10007;ll inv[N],fac[N],facInv[N];void getInv(int n){    inv[1]=fac[0]=facInv[0]=1;    for(int i=1;i<=n;i++){        if(i!=1) inv[i]=-P/i*inv[P%i]%P;           inv[i]+=inv[i]<0?P:0;        fac[i]=fac[i-1]*i%P;        facInv[i]=facInv[i-1]*inv[i]%P;    }}ll C(ll n,ll m){    if(n