zoukankan      html  css  js  c++  java
  • $\text{O}(n^2)$ 多项式运算

    namespace Math{
      
    #define MAXN 1001001
      
        int frac[MAXN],invf[MAXN],inv[MAXN];
      
        inline void init(int N){
            frac[0]=1;
            for(int i=1;i<=N;i++)
                frac[i]=1ll*frac[i-1]*i%mod;
            invf[N]=qpow(frac[N],mod-2);
            for(int i=N;i>=1;i--){
                invf[i-1]=1ll*invf[i]*i%mod;
                inv[i]=1ll*invf[i]*frac[i-1]%mod;
            }
        }
      
        inline long long C(int n,int m){
            return 1ll*frac[n]*invf[m]%mod*invf[n-m]%mod;
        }
      
    #undef MAXN
      
    }
      
    using Math::frac;
    using Math::invf;
    using Math::inv;
    
    template<int T>
    
    struct Polyn{
    
        int n,a[T];
    
        Polyn(int _n=0){
            n=_n;
            for(int i=0;i<n;i++)a[i]=0;
        }
    
        int& operator [](const int x){
            return a[x];
        }
    
        inline Polyn R(){
            Polyn b(n);
            for(int i=0;i*2<n;i++)
                b[i]=a[n-1-i];
            return b;
        }
      
        inline Polyn Der(){
            Polyn b(n-1);
            for(int i=1;i<n;i++)
                b[i-1]=1ll*a[i]*i%mod;
            return b;
        }
    
        inline Polyn Int(){
            Polyn b(n+1);
            for(int i=1;i<=n;i++)
                b[i]=1ll*a[i-1]*inv[i]%mod;
            return b;
        }
    
        inline Polyn Inv(){
            Polyn b(n);
            b[0]=qpow(a[0],mod-2);
            long long inv=b[0];
            for(int i=1;i<n;i++){
                long long res=0;
                for(int j=0;j<i;j++)
                    (res+=1ll*b[j]*a[i-j])%=mod;
                b[i]=(mod-res)*inv%mod;
            }
            return b;
        }
    
        Polyn operator -(){
            Polyn b(*this);
            for(int i=0;i<n;i++)
                b[i]=(mod-b[i])%mod;
            return b;
        }
    
        Polyn operator +(int x){
            Polyn b(*this);
            (b[0]+=x)%=mod;
            return b;
        }
    
        Polyn operator -(int x){
            Polyn b(*this);
            (b[0]+=mod-x)%=mod;
            return b;
        }
    
        Polyn operator *(int x){
            Polyn b(*this);
            for(int i=0;i<n;i++)
                b[i]=1ll*b[i]*x%mod;
            return b;
        }
    
        Polyn operator +(Polyn b){
            Polyn c=b;
            c.n=max(n,b.n);
            for(int i=b.n;i<c.n;i++)c[i]=0;
            for(int i=0;i<n;i++)
                (c[i]+=a[i])%=mod;
            return c;
        }
    
        Polyn operator -(Polyn b){
            Polyn c=(*this);
            c.n=max(n,b.n);
            for(int i=n;i<c.n;i++)c[i]=0;
            for(int i=0;i<b.n;i++)
                (c[i]+=mod-b[i])%=mod;
            return c;
        }
    
        Polyn operator *(Polyn b){
            Polyn c(n+b.n-1);
            for(int i=0;i<n;i++)
                for(int j=0;j<b.n;j++)
                    (c[i+j]+=1ll*a[i]*b[j]%mod)%=mod;
            return c;
        }
    
        Polyn operator /(Polyn b){
            return (*this)*b.Inv();
        }
    
        Polyn operator %(Polyn b){
            Polyn r=(*this);
            for(int i=n-1;i>=b.n-1;i--){
                int inv=r[i]*qpow(b[b.n-1],mod-2);
                for(int j=i;j>=i-b.n+1;j--)
                    (r[j]+=mod-1ll*b[j-i+b.n-1]*inv%mod)%=mod;
            }
            r.n=min(n,b.n-1);
            return r;
        }
    
        Polyn ln(){
            Polyn b(n);
            for(int i=1;i<n;i++){
                for(int j=1;j<i;j++)
                    (b[i]+=1ll*b[j]*a[i-j]%mod*j%mod)%=mod;
                b[i]=(a[i]+mod-1ll*b[i]*inv[i]%mod)%mod;
            }
            return b;
        }
    
        Polyn exp(){
            Polyn b(n);
            b[0]=1;
            for(int i=1;i<n;i++){
                for(int j=1;j<=i;j++)
                    (b[i]+=1ll*b[i-j]*a[j]%mod*j%mod)%=mod;
                b[i]=1ll*b[i]*inv[i]%mod;
            }
            return b;
        }
    
        inline int Y(int x){
            int Sum=0;
            for(int i=n-1;~i;i--)
                Sum=(1ll*Sum*x+a[i])%mod;
            return Sum;
        }
    
    };
  • 相关阅读:
    Leetcode 121. Best Time to Buy and Sell Stock
    Leetcode 120. Triangle
    Leetcode 26. Remove Duplicates from Sorted Array
    Leetcode 767. Reorganize String
    Leetcode 6. ZigZag Conversion
    KMP HDU 1686 Oulipo
    多重背包 HDU 2844 Coins
    Line belt 三分嵌套
    三分板子 zoj 3203
    二分板子 poj 3122 pie
  • 原文地址:https://www.cnblogs.com/wyzwyz/p/15549075.html
Copyright © 2011-2022 走看看