====== 2020.06.01-2020.06.07 周报 ======

===== 团队训练 =====
本周暂无

===== _wzx27 =====
==== 1.补题 ====
补了一下之前比赛的题

[[http://acm.hdu.edu.cn/showproblem.php?pid=6589|HDU6589 Sequence]]

把数列 $a_i$ 写成其生成函数的形式 $f(x)=\sum a_ix^i$，每个操作 $k$ 相当于 $f(x)\cdot \sum x^{ik}$，由交换律知顺序不重要，所以可以统计每种操作的次数 $m_i$，最后有

$$f(x)\cdot (\sum x^{i})^{m_1} \cdot (\sum x^{2i})^{m_2} \cdot (\sum x^{3i})^{m_3}$$

其中$(\sum x^{ki})^m = \sum {i+m-1 \choose m-1}x^{ik}$

另外模数刚好是998244353，做三次NTT即可

<hidden>
<code cpp>
#include<bits/stdc++.h>
#define ll long long
#define pii_ pair<int,int>
#define mp_ make_pair
#define pb push_back
#define fi first
#define se second
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define show1(a) cout<<#a<<" = "<<a<<endl
#define show2(a,b) cout<<#a<<" = "<<a<<"; "<<#b<<" = "<<b<<endl
using namespace std;
const ll INF = 1LL<<60;
const int inf = 1<<30;
const int maxn = 2e6+5;
const ll M = 998244353;
inline void fastio() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);}
ll qpow(ll a,ll b) {ll s=1;while(b){if(b&1)s=(s*a)%M;a=(a*a)%M;b>>=1;}return s; }
int n,m,lim,l,r[maxn],cnt[maxn];
ll A[maxn],B[maxn],fac[maxn],inv[maxn];
int c[5];
void NTT(ll a[],int type)
{
    rep(i,1,lim-1) if(i<r[i]) swap(a[i],a[r[i]]);
    for(int mid=1;mid<lim;mid<<=1){
        ll wn = qpow(3,(M-1)/mid/2);
        if(type==-1) wn = qpow(wn,M-2);
        for(int R=mid<<1,j=0;j<lim;j+=R){
            ll w = 1;
            for(int k=0;k<mid;k++,w=w*wn%M){
                ll x=a[j+k],y=w*a[j+mid+k]%M;
                a[j+k] = (x+y)%M;
                a[j+mid+k] = (x-y+M)%M;
            }
        }
    }
    if(type==-1){
        ll INV = qpow(lim,M-2);
        rep(i,0,lim) a[i] = a[i]*INV%M;
    }
}
ll Comb(ll a,ll b) {return a<b?0:fac[a]*inv[b]%M*inv[a-b]%M;}
void init()
{
    fac[0] = 1;
    rep(i,1,2000000) fac[i]=fac[i-1]*i%M;
    inv[2000000] = qpow(fac[2000000],M-2);
    per(i,2000000-1,0) inv[i]=inv[i+1]*(i+1)%M;
}
int main()
{
    fastio();
    int _;
    init();
    for(cin>>_;_;_--){
        cin>>n>>m;
        memset(A,0,sizeof(A));
        memset(B,0,sizeof(B));
        rep(i,0,n-1) cin>>A[i];
        lim=1,l=0;
        while(lim<=2*n) lim<<=1,l++;
        rep(i,0,lim-1) r[i] = (r[i>>1]>>1) | ((i&1)<<(l-1));
        cnt[1]=cnt[2]=cnt[3]=0;
        while(m--) {int x;cin>>x;cnt[x]++;}
        rep(i,1,3)if(cnt[i]){
            memset(B,0,sizeof(B));
            for(int j=0;j*i<n;j++) B[i*j] = Comb(cnt[i]+j-1,j);
            NTT(A,1);NTT(B,1);
            rep(i,0,lim) A[i] = A[i]*B[i]%M;
            NTT(A,-1);
            rep(i,n+1,lim) A[i]=0;
        }
        ll ans = 0;
        rep(i,0,n-1) ans ^= ((i+1)*A[i]);
        cout<<ans<<endl;
    }
    return 0;
}
</code>
</hidden>

==== 2.整数分拆 ====

生成函数的基本应用，求 $x1+x2+x3\cdots x^k=n$ 的非负整数解的个数。

=== 2.1 有序分拆 ===
把每个数写成生成函数 $(1+x+x^2+\cdots)$ 的形式

那么分拆数就等于 $(1+x+x^2+\cdots)^k$ 的 $n$ 次项前面的系数${n+k-1 \choose k-1}$

这个证明可以通过$(1+x+x^2+\cdots)^k=\frac 1{(1-x)^k}$的逐项求导证明

=== 2.2 无序分拆 ===
无序的分拆容易想到朴素 $O(n^2)$ 的$\text{dp}$

记 $P(n)$ 为 $n$ 的无序拆分数如果用生成函数来做就是

$$\sum_{i=1}^\infty P(i)x^i=(1+x+x^2+\cdots)(1+x^2+x^4+\cdots)\cdots=\prod_{i=0}^\infty \frac 1{1-x^i}$$

结合五边形数定理 $\prod_{i=0}^\infty (1-x^i)=1+\sum_{i=1}^\infty (-1)^i(x^{\frac {i(3i-1)}2}+x^{\frac {i(3i+1)}2})$ 得到：

$$(1+P(1)x+P(2)x^2+\cdots)(1-x-x^2+x^5+\cdots)=1$$

比较两边的系数得到 $P(n)-P(n-1)-P(n-2)+P(n-5)=0$，即$P(n)-P(n-1)-P(n-2)+P(n-5)+P(n-7)-\cdots = 0$

这样就得到了 $P(n)=\sum_{i=1}(-1)^i(P(n-\frac {i(3i-1)}2)+P(n-\frac {i(3i+1)}2))$

[[http://acm.hdu.edu.cn/showproblem.php?pid=4651|HDU4651]]
模板题
<hidden code>
<code cpp>
#include<bits/stdc++.h>
#define ll long long
#define pii_ pair<int,int>
#define mp_ make_pair
#define pb push_back
#define fi first
#define se second
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define show1(a) cout<<#a<<" = "<<a<<endl
#define show2(a,b) cout<<#a<<" = "<<a<<"; "<<#b<<" = "<<b<<endl
using namespace std;
const ll INF = 1LL<<60;
const int inf = 1<<30;
const int maxn = 2e5+5;
const ll M = 1e9+7;
inline void fastio() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);}

ll f1[maxn],f2[maxn],P[maxn];
void init()
{
    int n = 1e5;
    for(int i=1;;i++){
        f1[i] = (3*i*i-i)>>1;
        if(f1[i]>n) break;
        f2[i] = (3*i*i+i)>>1;
        if(f2[i]>n) break;
    }
    P[0] = 1;
    rep(i,1,n){
        for(int j=1;;j++){
            if(f1[j]<=i) P[i] += (j&1)?P[i-f1[j]] : -P[i-f1[j]];
            else break;
            if(f2[j]<=i) P[i] += (j&1)?P[i-f2[j]] : -P[i-f2[j]];
            else break;
        }
        P[i] = (P[i]%M+M)%M;
    }
}
int main()
{
    fastio();
    init();
    int _,n;
    for(cin>>_;_;_--){
        cin>>n;cout<<P[n]<<endl;
    }
    return 0;
}
</code>
</hidden>

[[http://acm.hdu.edu.cn/showproblem.php?pid=4658|HDU4658]]

拆分时每个数的使用次数要小于 $k$。

把生成函数改一下：

$$(1+x+x^2+\cdots+x^{k-1})(1+x^2+x^4+\cdots+x^{2(k-1)})\cdots=\prod_{i=1}^\infty \frac {1-x^{ik}}{1-x^i}=\phi(x^k)\sum_{i=0}^\infty P(n)x^i$$

<hidden>
<code cpp>
#include<bits/stdc++.h>
#define ll long long
#define pii_ pair<int,int>
#define mp_ make_pair
#define pb push_back
#define fi first
#define se second
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define show1(a) cout<<#a<<" = "<<a<<endl
#define show2(a,b) cout<<#a<<" = "<<a<<"; "<<#b<<" = "<<b<<endl
using namespace std;
const ll INF = 1LL<<60;
const int inf = 1<<30;
const int maxn = 2e5+5;
const ll M = 1e9+7;
inline void fastio() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);}

ll f1[maxn],f2[maxn],P[maxn];
void init()
{
    int n = 1e5;
    for(int i=1;;i++){
        f1[i] = (3*i*i-i)>>1;
        if(f1[i]>n) break;
        f2[i] = (3*i*i+i)>>1;
        if(f2[i]>n) break;
    }
    P[0] = 1;
    rep(i,1,n){
        for(int j=1;;j++){
            if(f1[j]<=i) P[i] += (j&1)?P[i-f1[j]] : -P[i-f1[j]];
            else break;
            if(f2[j]<=i) P[i] += (j&1)?P[i-f2[j]] : -P[i-f2[j]];
            else break;
        }
        P[i] = (P[i]%M+M)%M;
    }
}
ll solve(int n,int k)
{
    ll ans = P[n];
    for(int i=1;;i++){
        ll a = (ll)k*(3*i*i-i)>>1;
        if(a>n) break;
        ll t = P[n-a];
        ll b = (ll)k*(3*i*i+i)>>1;
        if(b<=n) t = (t+P[n-b])%M;
        if(i&1) ans = (ans-t+M)%M;
        else ans = (ans+t)%M;
    }
    return ans;
}
int main()
{
    fastio();
    init();
    int _,n,k;
    for(cin>>_;_;_--){
        cin>>n>>k;cout<<solve(n,k)<<endl;
    }
    return 0;
}
</code>
</hidden>

===== Infinity37 =====
在搞期末考试，完全摸了
===== Zars19 =====

不好意思摸了摸了，我马上补555

楼上上好强…

==== 比赛 ====

[[Educational Codeforces Round 83 (Rated for Div. 2)]]

===== 本周推荐 =====