BZOJ 4338 BJOI2015 糖果

此题其实没什么营养就提示了 ， 简单推一个公式 ， 再上中国剩余定理即可。

PnCmm+k−1

其中p指的排列公式。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <string>#include <vector>#include <deque>#include <stack>#include <queue>#include <set>#include <map>#include <algorithm>#include <cassert>using namespace std;typedef long long ll;ll  n , m , k , p , res;void exGcd(ll a , ll b , ll& d , ll& x , ll& y){    if(!b) x=1 , y=0 , d = a;    else     {        exGcd(b, a%b, d, y, x);        y -= a/b*x;    }}ll range(ll a , ll p) { return ((a%p)+p)%p; }ll inv(ll a , ll p){    ll d , x , y;    exGcd(a, p, d, x, y);    return range(x, p);}ll power(ll a , ll n){    ll res = 1;    while(n--) res*=a;    return res;}struct state{    ll t , a , p , rp;    state(ll p=0 , ll rp=0) { t = 0; a = 1; this->p = p; this->rp = rp; }    state operator *(ll a)    {        state res = *this;        while(a%p==0) res.t++ , a/=p;        a%=rp;        res.a = range(res.a*a, rp);        return res;    }    state operator /(ll a)    {        state res = *this;        while(a%p==0) res.t-- , a/=p;        a%=rp;        res.a = range(res.a*inv(a, rp), rp);        return res;    }};void solve(ll p , ll t){    state now = state(p , power(p, t));    for(ll i=1;i<=m;i++)     {        now = now*(m+k-i);        now = now/i;    }    ll b = range(now.a*power(now.p, now.t), now.rp) , res=1;    for(ll i=0;i<n;i++) res = range(res*(b-i), now.rp);    ll add = range(inv(::p/now.rp, now.rp)*(::p/now.rp), ::p);    add = range(add*res, ::p);    ::res = range(::res+add, ::p);}int main(int argc, char *argv[]) {    cin>>n>>m>>k>>p;    ll _p = p;    for(ll i=2;i*i<=p;i++)if(p%i==0)    {        ll t = 0;        while(_p%i==0) t++ , _p/=i;        solve(i, t);    }    if(_p!=1) solve(_p, 1);    cout<<res<<endl;    return 0;}