拉格朗日插值板子

左手的ㄟ右手 2021-09-19 18:40 318阅读 0赞

其实这明明是个polysum的板子，为什么都叫它插值呢？？？

By 杜教

namespace polysum { 
        #define rep(i,a,n) for (int i=a;i<n;i++)
        #define per(i,a,n) for (int i=n-1;i>=a;i--)
        const int D=2010;
        ll a[D],f[D],g[D],p[D],p1[D],p2[D],b[D],h[D][2],C[D];
        ll powmod(ll a,ll b){ ll res=1;a%=mod;assert(b>=0);for(;b;b>>=1){ if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
        ll calcn(int d,ll *a,ll n) {  // a[0].. a[d] a[n] 
            if (n<=d) return a[n];
            p1[0]=p2[0]=1;
            rep(i,0,d+1) { 
                ll t=(n-i+mod)%mod;
                p1[i+1]=p1[i]*t%mod;
            }
            rep(i,0,d+1) { 
                ll t=(n-d+i+mod)%mod;
                p2[i+1]=p2[i]*t%mod;
            }
            ll ans=0;
            rep(i,0,d+1) { 
                ll t=g[i]*g[d-i]%mod*p1[i]%mod*p2[d-i]%mod*a[i]%mod;
                if ((d-i)&1) ans=(ans-t+mod)%mod;
                else ans=(ans+t)%mod;
            }
            return ans;
        }
        void init(int M) { 
            f[0]=f[1]=g[0]=g[1]=1;
            rep(i,2,M+5) f[i]=f[i-1]*i%mod;
            g[M+4]=powmod(f[M+4],mod-2);
            per(i,1,M+4) g[i]=g[i+1]*(i+1)%mod;
        }
        ll polysum(ll m,ll *a,ll n) {  // a[0].. a[m] \sum_{i=0}^{n-1} a[i]
            ll b[D];
            for(int i=0;i<=m;i++) b[i]=a[i];
            b[m+1]=calcn(m,b,m+1);
            rep(i,1,m+2) b[i]=(b[i-1]+b[i])%mod;
            return calcn(m+1,b,n-1);
        }
        ll qpolysum(ll R,ll n,ll *a,ll m) {  // a[0].. a[m] \sum_{i=0}^{n-1} a[i]*R^i
            if (R==1) return polysum(n,a,m);
            a[m+1]=calcn(m,a,m+1);
            ll r=powmod(R,mod-2),p3=0,p4=0,c,ans;
            h[0][0]=0;h[0][1]=1;
            rep(i,1,m+2) { 
                h[i][0]=(h[i-1][0]+a[i-1])*r%mod;
                h[i][1]=h[i-1][1]*r%mod;
            }
            rep(i,0,m+2) { 
                ll t=g[i]*g[m+1-i]%mod;
                if (i&1) p3=((p3-h[i][0]*t)%mod+mod)%mod,p4=((p4-h[i][1]*t)%mod+mod)%mod;
                else p3=(p3+h[i][0]*t)%mod,p4=(p4+h[i][1]*t)%mod;
            }
            c=powmod(p4,mod-2)*(mod-p3)%mod;
            rep(i,0,m+2) h[i][0]=(h[i][0]+h[i][1]*c)%mod;
            rep(i,0,m+2) C[i]=h[i][0];
            ans=(calcn(m,C,n)*powmod(R,n)-c)%mod;
            if (ans<0) ans+=mod;
            return ans;
        }
    }