题意:n个物品每个价值a[i],要求选k个,可以重复,问能取到哪几个价值

题解:fft裸题.但是直接一次fft,然后快速幂会boom.这样是严格的\(2^{20}*log2(2^{20})*log(w)\).需要在快速幂里fft,每次取最大的2的次幂,然后fft也boom了,不知道是不是写搓了.ntt过了.....

//#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>

#define fi first

#define se second

#define db double

#define mp make_pair

#define pb push_back

#define pi acos(-1.0)

#define ll long long

#define vi vector<int>

#define mod 998244353

#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 base 1000000000000000000

#define fin freopen("a.txt","r",stdin)

#define fout freopen("a.txt","w",stdout)

#define fio ios::sync_with_stdio(false);cin.tie(0)

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 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=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const ull ba=233;

const db eps=1e-5;

const ll INF=0x3f3f3f3f3f3f3f3f;

const int N=1000000+10,maxn=1000000+10,inf=0x3f3f3f3f;

ll x[N<<3],y[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 ntt(ll *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)

    {

        ll wn=qp(3,(mod-1)/(step*2));

        if(dft==-1)wn=qp(wn,mod-2);

        for(int j=0;j<n;j+=step<<1)

        {

            ll wnk=1;

            for(int k=j;k<j+step;k++)

            {

                ll x=a[k];

                ll y=wnk*a[k+step]%mod;

                a[k]=(x+y)%mod;a[k+step]=(x-y+mod)%mod;

                wnk=wnk*wn%mod;

            }

        }

    }

    if(dft==-1)

    {

        ll inv=qp(n,mod-2);

        for(int i=0;i<n;i++)a[i]=a[i]*inv%mod;

    }

}

void solve(int k,int p)

{

    int sz=0;while((1<<sz)<=p)sz++;sz++;

    getrev(sz);

    ntt(y,(1<<sz),1);

    if(k&1)

    {

        ntt(x,(1<<sz),1);

        for(int i=0;i<(1<<sz);i++)x[i]=x[i]*y[i]%mod;

        ntt(x,(1<<sz),-1);

        for(int i=0;i<(1<<sz);i++)if(x[i])x[i]=1;

    }

    for(int i=0;i<(1<<sz);i++)y[i]=y[i]*y[i]%mod;

    ntt(y,(1<<sz),-1);

    for(int i=0;i<(1<<sz);i++)if(y[i])y[i]=1;

}

int main()

{

    int n,k,ma=0;scanf("%d%d",&n,&k);

    for(int i=1;i<=n;i++)

    {

        int a;scanf("%d",&a);

        y[a]+=1;ma=max(ma,a);

    }

    x[0]=1;

    while(k)solve(k,ma),k>>=1,ma<<=1;

    for(int i=1;i<N;i++)if(x[i]!=0)printf("%d ",i);

    return 0;

}

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

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