• luogu 4725 【模板】多项式对数函数(多项式 ln)


    $G(x)=ln(A(x))$

    $G'(x)=ln'(A(x))A'(x)=frac{A'(x)}{A(x)}$    

    由于求导和积分是互逆的,所以对 $G$ 求积分,即

    $G(x)=intfrac{A'(x)}{A(x)}$

    用求导 + 求逆 + 积分做一下即可

    这里给出求导/积分的公式:

    $int F(x)=sum_{i=0}^{n}frac{a_{i}}{i+1}x^{i+1}$

    $d(F(x))=sum_{i=1}^{n}i imes a_{i}x^{i-1}$         
    #include <cstdio>
    #include <string>
    #include <algorithm>    
    #include <cstring>
    #include <vector>     
    #define setIO(s) freopen(s".in","r",stdin)   
    typedef long long ll;
    const int maxn=2100005;  
    const ll mod=998244353; 
    using namespace std;                   
    inline ll qpow(ll base,ll k) {
        ll tmp=1;      
        for(;k;k>>=1,base=base*base%mod)if(k&1) tmp=tmp*base%mod;   
        return tmp;     
    }       
    inline ll inv(ll a) { return qpow(a, mod-2); }     
    inline void NTT(ll *a,int len,int flag) {
        for(int i=0,k=0;i<len;++i) {
            if(i>k) swap(a[i],a[k]);     
            for(int j=len>>1;(k^=j)<j;j>>=1);  
        }  
        for(int mid=1;mid<len;mid<<=1) {
            ll wn=qpow(3, (mod-1)/(mid<<1)),x,y;  
            if(flag==-1) wn=qpow(wn,mod-2);  
            for(int i=0;i<len;i+=(mid<<1)) {     
                ll w=1;
                for(int j=0;j<mid;++j) {        
                    x=a[i+j],y=w*a[i+j+mid]%mod;   
                    a[i+j]=(x+y)%mod, a[i+j+mid]=(x-y+mod)%mod;   
                    w=w*wn%mod;    
                }
            }  
        }
        if(flag==-1) {
            int re=qpow(len,mod-2); 
            for(int i=0;i<len;++i) a[i]=a[i]*re%mod;   
        }
    }
    ll A[maxn],B[maxn];           
    struct poly {
        vector<ll>a; 
        int len; 
        poly(){}                
        inline void clear() { len=0; a.clear(); }   
        inline void rev() {reverse(a.begin(), a.end()); }              
        inline void push(int x) { a.push_back(x),++len; }   
        inline void resize(int x) { len=x; a.resize(x); }                  
        void getinv(poly &b,int n) {
            if(n==1) { b.clear(); b.push(inv(a[0]));  return; } 
            getinv(b,n>>1);     
            int t=n<<1,lim=min(len,n);
            for(int i=0;i<lim;++i) A[i]=a[i];
            for(int i=lim;i<t;++i) A[i]=0;
            for(int i=0;i<b.len;++i) B[i]=b.a[i];
            for(int i=b.len;i<t;++i) B[i]=0;  
            NTT(A,t,1),NTT(B,t,1);  
            for(int i=0;i<t;++i)  A[i]=(2-A[i]*B[i]%mod+mod)*B[i]%mod; 
            NTT(A,t,-1);             
            b.clear();          
            for(int i=0;i<n;++i) b.push(A[i]);   
        }    
        poly Inv() {
            int n=1;
            while(n<=len)n<<=1;  
            poly b;           
            b.clear(), getinv(b,n);          
            return b;                         
        }         
        poly dao() {
            poly c;    
            c.resize(len);        
            for(int i=1;i<=len;++i) c.a[i-1]=a[i]*i%mod; 
            return c; 
        }  
        poly jifen() {
            poly c; 
            c.resize(len+1);   
            for(int i=1;i<=len;++i) c.a[i]=a[i-1]*qpow(i,mod-2)%mod; 
            c.a[0]=0; 
            return c;   
        }
        poly Ln() {
            poly c=dao()*Inv();
            return c.jifen();    
        }
        poly operator * (const poly &b) const {
            int n=1;
            while(n<=len+b.len) n<<=1;  
            for(int i=0;i<len;++i) A[i]=a[i];
            for(int i=len;i<n;++i) A[i]=0;
            for(int i=0;i<b.len;++i) B[i]=b.a[i];
            for(int i=b.len;i<n;++i) B[i]=0;
            NTT(A,n,1), NTT(B,n,1);
            for(int i=0;i<n;++i) A[i]=A[i]*B[i]%mod;    
            NTT(A,n,-1);    
            poly c;
            c.clear();
            for(int i=0;i<len+b.len-1;++i) c.push(A[i]);    
            return c;        
        }     
        poly operator + (const poly &b) const {
            poly c; 
            c.clear();    
            for(int i=0;i<len;++i) c.push(a[i]); 
            for(int i=0;i<b.len;++i) {
                if(i<len) c.a[i]=(c.a[i]+b.a[i])%mod;    
                else c.push(b.a[i]); 
            }
            return c;     
        }
        poly operator - (const poly &b) const {
            poly c; 
            c.clear();   
            for(int i=0;i<len;++i) c.push(a[i]); 
            for(int i=0;i<b.len;++i) {
                if(i<len) c.a[i]=(c.a[i]-b.a[i]+mod)%mod;  
                else c.push((mod-b.a[i])%mod);  
            }
            return c;  
        }
        friend poly operator / (poly f,poly g) {   
            poly Q;       
            int l=f.len-g.len+1;
            f.rev(), g.rev(), g.resize(l), f.resize(l);                  
            g=g.Inv(), Q=f*g, Q.resize(l),Q.rev();           
            return Q;   
        }
        friend poly operator % (poly f,poly g) {
            poly u=f-(f/g)*g;    
            u.resize(g.len-1);
            return u;     
        }              
    }po[4];        
    inline void inv() {
        int n,x;
        scanf("%d",&n), po[0].clear();
        for(int i=0;i<n;++i) scanf("%d",&x), po[0].push(x);   
        po[1]=po[0].Inv();
        for(int i=0;i<po[1].len;++i) printf("%lld ",po[1].a[i]);  
    }         
    inline void mult() {
        int n,m,x;
        scanf("%d%d",&n,&m);
        for(int i=0;i<=n;++i) scanf("%d",&x), po[0].push(x);
        for(int i=0;i<=m;++i) scanf("%d",&x), po[1].push(x);   
        po[1]=po[0]*po[1]; 
        for(int i=0;i<po[1].len;++i) printf("%lld ",po[1].a[i]);
    }                           
    inline void divide() {
        int n,m,x;
        scanf("%d%d",&n,&m);                           
        for(int i=0;i<=n;++i) scanf("%d",&x), po[0].push(x);   
        for(int i=0;i<=m;++i) scanf("%d",&x), po[1].push(x);  
        po[2]=po[0]/po[1];           
        for(int i=0;i<po[2].len;++i) printf("%lld ",po[2].a[i]); 
        printf("
    "); 
        po[2]=po[0]%po[1];
        for(int i=0;i<po[2].len;++i) printf("%lld ",po[2].a[i]); 
    }
    inline void Ln() {
        int n,x;
        scanf("%d",&n); 
        for(int i=0;i<n;++i) scanf("%d",&x), po[0].push(x); 
        po[0]=po[0].Ln();  
        for(int i=0;i<n;++i) printf("%lld ",po[0].a[i]);  
    }
    int main() {
        // setIO("input");  
        Ln(); 
        return 0; 
    }
    

      

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  • 原文地址:https://www.cnblogs.com/guangheli/p/11291615.html
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