Helvetic Coding Contest 2018 online mirror (teams allowed, unrated)F3 - Lightsabers (hard)

题意:n个数字1-m,问取k个组成的set方案数

题解:假设某个数出现k次,那么生成函数为\(1+x+...+x^k\),那么假设第i个数出现ai次,结果就是\(\sum_{i=1}^m(1+x+...+x^{a_i})\),第k项即为答案,启发式合并fft即可

组合(即set):普通生成函数.排列:指数型生成函数

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
//#include <bits/extc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define mt make_tuple
//#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1009
#define ld long double
//#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define ull unsigned long long
#define bpc __builtin_popcount
#define base 1000000000000000000ll
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
#define mr mt19937 rng(chrono::steady_clock::now().time_since_epoch().count())
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll mul(ll a,ll b,ll c){return (a*b-(ll)((ld)a*b/c)*c+c)%c;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=mul(ans,a,c);a=mul(a,a,c),b>>=1;}return ans;}

using namespace std;
//using namespace __gnu_pbds;

const ld pi = acos(-1);
const ull ba=233;
const db eps=1e-5;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=200000+10,maxn=2000000+10,inf=0x3f3f3f3f;

struct cd{
    ld x,y;
    cd(ld _x=0.0,ld _y=0.0):x(_x),y(_y){}
    cd operator +(const cd &b)const{
        return cd(x+b.x,y+b.y);
    }
    cd operator -(const cd &b)const{
        return cd(x-b.x,y-b.y);
    }
    cd operator *(const cd &b)const{
        return cd(x*b.x - y*b.y,x*b.y + y*b.x);
    }
    cd operator /(const db &b)const{
        return cd(x/b,y/b);
    }
}a[N<<3],b[N<<3];
int rev[N<<3];
void getrev(int bit)
{
    for(int i=0;i<(1<<bit);i++)
        rev[i]=(rev[i>>1]>>1) | ((i&1)<<(bit-1));
}
void fft(cd *a,int n,int dft)
{
    for(int i=0;i<n;i++)
        if(i<rev[i])
            swap(a[i],a[rev[i]]);
    for(int step=1;step<n;step<<=1)
    {
        cd wn(cos(dft*pi/step),sin(dft*pi/step));
        for(int j=0;j<n;j+=step<<1)
        {
            cd wnk(1,0);
            for(int k=j;k<j+step;k++)
            {
                cd x=a[k];
                cd y=wnk*a[k+step];
                a[k]=x+y;a[k+step]=x-y;
                wnk=wnk*wn;
            }
        }
    }
    if(dft==-1)for(int i=0;i<n;i++)a[i]=a[i]/n;
}
int c[N];
struct info{
    vi v;
    int id;
    bool operator <(const info&rhs)const{
        return v.size()<rhs.v.size()||(v.size()==rhs.v.size()&&id<rhs.id);
    }
};
set<info>s;
int main()
{
    int n,m,k;scanf("%d%d%d",&n,&m,&k);
    for(int i=1,x;i<=n;i++)scanf("%d",&x),c[x]++;
    for(int i=1;i<=m;i++)if(c[i])
    {
        info te;te.v.clear();te.id=i;
        for(int j=0;j<=c[i];j++)te.v.pb(1);
        s.insert(te);
    }
    while(s.size()>=2)
    {
        info x=*s.begin();s.erase(s.begin());
        info y=*s.begin();s.erase(s.begin());
        int p=x.v.size(),q=y.v.size();
        int sz=0;
        while((1<<sz)<=p+q)sz++;
        getrev(sz);
        int len=(1<<sz);
        for(int i=0;i<len;i++)
        {
            a[i]=(i<p?x.v[i]:0);
            b[i]=(i<q?y.v[i]:0);
        }
        fft(a,len,1);fft(b,len,1);
        for(int i=0;i<len;i++)a[i]=a[i]*b[i];
        fft(a,len,-1);
        x.v.clear();
        for(int i=0;i<len;i++)x.v.pb(((ll)(a[i].x+0.5))%mod);
        while(x.v.size()&&x.v.back()==0)x.v.pop_back();
        s.insert(x);
    }
    printf("%d\n",s.begin()->v[k]);
    return 0;
}
/********************

********************/