Bubble Cup 12 - Finals [Online Mirror, unrated, Div. 1] E. Product Tuples

题意略,题解生成函数练习题,1+(q-ai)x卷积即可,线段树优化(类似分治思想)
//#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 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 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;

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;

    }

}

vi v[N<<2];

int a[N],b[N];

void solve(int l,int r,int rt)

{

    if(l==r)

    {

        v[rt].clear();

        v[rt].pb(1);v[rt].pb(b[l]);

//        printf("%d\n",b[l]);

        return ;

    }

    int m=(l+r)>>1;

    solve(ls);solve(rs);

    int p=v[rt<<1].size(),q=v[rt<<1|1].size();

    int sz=0;

    while((1<<sz)<=p+q)sz++;

    getrev(sz);

    int len=(1<<sz);

//    printf("%d\n",len);

    for(int i=0;i<len;i++)

    {

        x[i]=(i<p?v[rt<<1][i]:0);

        y[i]=(i<q?v[rt<<1|1][i]:0);

    }

    ntt(x,len,1);ntt(y,len,1);

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

    ntt(x,len,-1);

    v[rt].clear();

    for(int i=0;i<len;i++)v[rt].pb(x[i]);

    while(v[rt].size()&&v[rt].back()==0)v[rt].pop_back();

}

int main()

{

//    fin;

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

    for(int i=1;i<=n;i++)scanf("%d",&a[i]),a[i]%=mod;

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

    while(q--)

    {

        int op,q,x,y,z;scanf("%d%d%d%d",&op,&q,&x,&y);

        if(op==1)

        {

            q%=mod,y%=mod;

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

            {

                b[i]=(i==x?q-y:q-a[i]);

                if(b[i]<0)b[i]+=mod;

            }

        }

        else

        {

            scanf("%d",&z);

            q%=mod,z%=mod;

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

            {

                if(x<=i&&i<=y)b[i]=((q-a[i]-z)%mod+mod)%mod;

                else b[i]=(q-a[i]+mod)%mod;

            }

        }

        solve(1,n,1);

        printf("%d\n",v[1][k]);

    }

    return 0;

}

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

********************/
巴特西

Bubble Cup 12 - Finals [Online Mirror, unrated, Div. 1] E. Product Tuples

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