CF632E： Thief in a Shop（快速冪+NTT）（存疑）

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A thief made his way to a shop.

As usual he has his lucky knapsack with him. The knapsack can contain k objects. There are n kinds of products in the shop and an infinite number of products of each kind. The cost of one product of kind i is a_i.

The thief is greedy, so he will take exactly k products (it‘s possible for some kinds to take several products of that kind).

Find all the possible total costs of products the thief can nick into his knapsack.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 1000) — the number of kinds of products and the number of products the thief will take.

The second line contains n integers a_{i (1 ≤ ai ≤ 1000) — the costs of products for kinds from 1 to n.}

Output

Print the only line with all the possible total costs of stolen products, separated by a space. The numbers should be printed in the ascending order.

Examples

Input

3 2
1 2 3

Output

2 3 4 5 6

Input

5 5
1 1 1 1 1

Output

5

Input

3 3
3 5 11

Output

9 11 13 15 17 19 21 25 27 33

題意：給定N個數a[]，讓你選擇K個數，可以重複選，求其組合成的和有哪些。N、K、a[]<=1000;

思路：看成1000000項的多項式，如果存在a[]=x，則x的係數為1，然後多項式自乘K次，係數不為0的部分表示可以有K個數構成，可以用FFT+快速冪，為了避免精度誤差，每次快速冪後把非0的改為1，免得變得很大後產生誤差，複雜度O(1000000*log1000000*logK)，有點大，稍微最佳化下常數可以卡過。

這裡嘗試 用NTT，由於係數可以達到1000^1000，所以需要除Mod，但是避免除一個Mod恰好變為0，所以我們取兩個Mod避免hack。

快速冪+NTT   4398ms：

#include<bits/stdc++.h>#define rep(i,x,y) for(int i=x;i<=y;i++)using namespace std;#define MOD Mod#define ll long longconst int G=3;const int maxn=5685760;int Mod;int qpow(int v,int p){    int ans=1;    for(;p;p>>=1,v=1ll*v*v%Mod)      if(p&1)ans=1ll*ans*v%Mod;    return ans;}void rader(int y[], int len) {    for(int i=1,j=len/2;i<len-1;i++) {        if(i<j) swap(y[i],y[j]);        int k=len/2;        while(j>=k) j-=k,k/=2;        if(j<k) j+=k;    }}void NTT(int y[],int len,int opt) {    rader(y,len);    for(int h=2;h<=len;h<<=1) {        int wn=qpow(G,(MOD-1)/h);        if(opt==-1) wn=qpow(wn,Mod-2);        for(int j=0;j<len;j+=h) {            int w=1;            for(int k=j;k<j+h/2;k++) {                int u=y[k];                int t=(ll)w*y[k+h/2]%MOD;                y[k]=(u+t)%MOD;                y[k+h/2]=(u-t+MOD)%MOD;                w=(ll)w*wn%MOD;            }        }    }    if(opt==-1) {        int t=qpow(len,MOD-2);        for(int i=0;i<len;i++) y[i]=(ll)y[i]*t%MOD;    }}void powNTT(int ans[],int a[],int x){    ans[0]=1;int len=1024;    while(x){        len<<=1;        if(x&1){            NTT(ans,len,1); NTT(a,len,1);            rep(i,0,len-1) ans[i]=(ll)ans[i]*a[i]%Mod;            NTT(ans,len,-1); NTT(a,len,-1);        }        NTT(a,len,1);        rep(i,0,len-1) a[i]=(ll)a[i]*a[i]%Mod;        NTT(a,len,-1);        x>>=1;    }}int A[maxn],B[maxn],ans1[maxn],ans2[maxn];int main(){    int N,K,x;    scanf("%d%d",&N,&K);    rep(i,1,N) scanf("%d",&x),A[x]=1,B[x]=1;    Mod=998244353; powNTT(ans1,A,K);    Mod=1004535809; powNTT(ans2,B,K);    rep(i,1,1000000) if(ans1[i]||ans2[i]) printf("%d ",i);    return 0;}

洛穀給出的代碼，https://www.luogu.org/problemnew/solution/CF632E ，只一次NTT，在DFT後把每個數單自求pow(K)，就得到了正確答案。

（暫時不理解其解法的正確性，如果是正確的，其NTT的寫法裡可能也有玄機（因為把這個NTT板子套其他題，範例過不了），尚待解決。

#include<bits/stdc++.h>#define rep(i,a,b) for(int i=a;i<=b;i++)using namespace std;#define ll long longconst int G=3;const int maxn=1048576;int mod,n,k,rev[maxn],lim,ilim,s,wn[maxn+1];std::vector<int> v;inline int pow(int x, int y) {    int ans=1;    for(;y;y>>=1,x=(ll)x*x%mod)      if(y&1) ans=(ll)ans*x%mod;    return ans;}inline int& up(int& x, int y) { if ((x+=y)>=mod) x-=mod; return x; }inline void NTT(int* A, int typ) {    rep(i,0,lim-1) if (i<rev[i]) swap(A[i], A[rev[i]]);    for (int i=1;i<lim;i+=i) {        const int t=lim/i/2;        for (int j=0;j<lim;j+=i+i) {            for (int k=0;k<i; k++) {                int w=typ?wn[t*k]:wn[lim-t*k];                int x=A[k+j],y=(ll)w*A[k+j+i]%mod;                up(A[k+j],y),up(A[k+j+i]=x,mod-y);            }        }    }    if (!typ) rep(i,0,lim-1) A[i]=(ll)ilim*A[i]%mod;}inline void init(int len,int tmod) {    mod=tmod; lim=1; s=-1;    while(lim<len) lim+=lim,s++;  ilim=pow(lim,mod-2);    rep(i,0,lim-1) rev[i]=rev[i>>1]>>1|(i&1)<<s;    int w=pow(G,(mod-1)/len);    wn[0]=1;    rep(i,1,lim) wn[i]=(ll)(wn[i-1])*w%mod;}int A[maxn], B[maxn];int main() {    scanf("%d%d",&n,&k);    int x; rep(i,1,n) scanf("%d",&x), A[x]=B[x]=1;    init(1048576, 998244353);    NTT(A, 1);    rep(i,0,lim-1) A[i]=pow(A[i],k);    NTT(A, 0);    rep(i,1,1000000) if (A[i]) v.push_back(i);    init(1048576, 1004535809);    NTT(B, 1);    for (int i = 0; i < lim; i++) B[i] = pow(B[i], k);    NTT(B, 0);    rep(i,1,1000000) if (B[i]) v.push_back(i);    sort(v.begin(), v.end()); int tot=unique(v.begin(), v.end())-v.begin();    v.resize(tot);    for (int i : v) printf("%d ",i);    return 0;}

 

CF632E： Thief in a Shop（快速冪+NTT）（存疑）