[BJOI2019]勘破神机

淡淡的烟草味﹌ 2021-12-23 15:07 248阅读 0赞

[\[BJOI2019\]勘破神机][BJOI2019]

推式子好题

m=2，斐波那契数列，$f\_\{n+1\}$项

不妨$++l,++r$，直接求$f\_n$

求$\\sum C(f\_n,k)$，下降幂转化成阶乘幂，这样都是多项式了，方便交换求和号

最后面的斐波那契数列用通项公式求。二项式展开。

交换求和号之后，枚举i,j 最后一项是等比数列求和。

[%rqy][rqy]

![1399631-20190425090447726-16565619.png][]

m=3，

n为奇数是0

n是偶数时，令n=n/2 递推公式：$g\_n=4\\times g\_\{n-1\}+g\_\{n-2\}$

证明：枚举从后往前第一个完全分出的块，除了块长为2的方案额外多一个外，其它都是两种。$g\_n=g\_\{n-1\}+2\\times \\sum\_\{i=0\}^\{n-1\} g\_\{i\}$

再写出：$g\_\{n-1\}=g\_\{n-2\}+2\\times \\sum\_\{i=0\}^\{n-2\} g\_\{i\}$两式做差移项即可得到。

用特征方程可以解得$g\_n$的通项公式

![1399631-20190425090519182-1233896492.png][]

$\\sqrt 5$在mod 998244353下不存在，可以用$a+b\\sqrt5$形式表示

注意，等比数列求和：$1+Q+....+Q^n=\\frac\{1-Q^\{n+1\}\}\{1-Q\}$注意是n+1，因为有n+1项

#include<bits/stdc++.h>
    #define reg register int
    #define il inline
    #define fi first
    #define se second
    #define mk(a,b) make_pair(a,b)
    #define numb (ch^'0')
    using namespace std;
    typedef long long ll;
    template<class T>il void rd(T &x){
        char ch;x=0;bool fl=false;
        while(!isdigit(ch=getchar()))(ch=='-')&&(fl=true);
        for(x=numb;isdigit(ch=getchar());x=x*10+numb);
        (fl==true)&&(x=-x);
    }
    template<class T>il void output(T x){
        if(x/10)output(x/10);putchar(x%10+'0');}
    template<class T>il void ot(T x){
        if(x<0) putchar('-'),x=-x;output(x);putchar(' ');}
    template<class T>il void prt(T a[],int st,int nd){
        for(reg i=st;i<=nd;++i) ot(a[i]);putchar('\n');}
    
    namespace Miracle{
    const int mod=998244353;
    const int N=505;
    int C;
    int ad(int x,int y){
        return x+y>=mod?x+y-mod:x+y;
    }
    int mul(int x,int y){
        return (ll)x*y%mod;
    }
    ll qm(ll x,ll y){
        ll ret=1;
        while(y){
            if(y&1) ret=mul(ret,x);
            x=mul(x,x);
            y>>=1;
        }
        return ret;
    }
    int inv[1000000+5];
    int ni(int x){
        // cout<<" ni x "<<x<<endl;
        return x<=1000000?inv[x]:qm(x,mod-2);
    }
    struct po{
        int a,b;
        po(){
            a=0;b=0;
        }
        po(int aa,int bb){
            a=aa;b=bb;
        }
        po friend operator +(po a,po b){
            return po(ad(a.a,b.a),ad(a.b,b.b));
        }
        po friend operator -(po a,po b){
            return po(ad(a.a,mod-b.a),ad(a.b,mod-b.b));
        }
        po friend operator ~(po a){
            int mom=ni(ad(mul(a.a,a.a),mod-mul(C,mul(a.b,a.b))));
            // cout<<" mom "<<mom<<endl;
            return po(mul(a.a,mom),ad(0,mod-mul(a.b,mom)));
        }
        po friend operator -(po a){
            return po(ad(0,mod-a.a),ad(0,mod-a.b));
        }
        po friend operator *(po a,po b){
            return po(ad(mul(a.a,b.a),mul(mul(a.b,b.b),C)),ad(mul(a.a,b.b),mul(a.b,b.a)));
        }
        po friend operator *(po a,int c){
            return po(mul(a.a,c),mul(a.b,c));
        }
        po friend operator /(po a,po b){
            return a*(~b);
        }
        void op(){
            cout<<" a "<<a<<" b "<<b<<endl;
        }
    }A,B,X,Y,mi[N][4];
    po qm(po x,ll y){
        po ret;ret.a=1;
        while(y){
            if(y&1) ret=ret*x;
            x=x*x;
            y>>=1;
        }
        return ret;
    }
    po calc(po Q,ll n){
        // Q.op();
        if(Q.a==1&&Q.b==0){
            return po((n+1)%mod,0);
        }
        po tmp=Q;tmp=qm(tmp,n+1);
        tmp=-tmp;tmp.a=ad(tmp.a,1);
        Q=-Q;Q.a=ad(Q.a,1);
        // Q.op();
        // Q=~Q;
        // Q.op();
        return tmp*(~Q);
    }
    int s[N][N],c[N][N];
    int main(){
        int t;rd(t);int m;rd(m);
        inv[1]=1;
        for(reg i=2;i<=1000000;++i){
            inv[i]=mul(mod-mod/i,inv[mod%i]);
        }
        if(m==2) {
            C=5;A=po(0,ni(5));B=po(0,mod-ni(5));
            X=po(ni(2),ni(2));Y=po(ni(2),mod-ni(2));
        }
        else {
            C=3;A=po(ni(2),ni(6));B=po(ni(2),mod-ni(6));
            X=po(2,1);Y=po(2,mod-1);
        }
        mi[0][0]=mi[0][1]=mi[0][2]=mi[0][3]=po(1,0);
        for(reg i=1;i<=502;++i){
            mi[i][0]=mi[i-1][0]*A;
            mi[i][1]=mi[i-1][1]*B;
            mi[i][2]=mi[i-1][2]*X;
            mi[i][3]=mi[i-1][3]*Y;
        }
    
        s[0][0]=1;
        for(reg i=1;i<=502;++i){
            for(reg j=1;j<=502;++j){
                s[i][j]=ad(mul(s[i-1][j],i-1),s[i-1][j-1]);
            }
        }
        c[0][0]=1;
        for(reg i=1;i<=502;++i){
            c[i][0]=1;
            for(reg j=1;j<=502;++j){
                c[i][j]=ad(c[i-1][j-1],c[i-1][j]);
            }
        }
    
        ll l,r,k;
        while(t--){
            rd(l);rd(r);rd(k);
            if(m==2) {
                ++l,++r;
                po ans;
                for(reg i=0;i<=k;++i){
                    // cout<<" i "<<i<<endl;
                    po tmp;
                    for(reg j=0;j<=i;++j){
                        // cout<<" jj "<<j<<endl;
                        tmp=tmp+mi[j][0]*mi[i-j][1]*(calc(mi[j][2]*mi[i-j][3],r)-calc(mi[j][2]*mi[i-j][3],l-1))*c[i][j];
                        // cout<<" bac "<<endl;
                    }
                    if((k-i)&1) tmp=-tmp;
                    ans=ans+(tmp*s[k][i]);
                }
                for(reg i=1;i<=k;++i) ans=ans*inv[i];
                ans=ans*qm((r-l+1)%mod,mod-2);
                
                printf("%d\n",ans.a);
            }
            else{ 
                ll L=l,R=r;
                l=(l+1)/2,r=r/2;
                po ans;
                for(reg i=0;i<=k;++i){
                    // cout<<" i "<<i<<endl;
                    po tmp;
                    for(reg j=0;j<=i;++j){
                        // cout<<" jj "<<j<<endl;
                        tmp=tmp+mi[j][0]*mi[i-j][1]*(calc(mi[j][2]*mi[i-j][3],r)-calc(mi[j][2]*mi[i-j][3],l-1))*c[i][j];
                        // cout<<" bac "<<endl;
                    }
                    if((k-i)&1) tmp=-tmp;
                    ans=ans+(tmp*s[k][i]);
                }
                for(reg i=1;i<=k;++i) ans=ans*inv[i];
                ans=ans*qm((R-L+1)%mod,mod-2);
    
                // ans.op();
                printf("%d\n",ans.a);
            }
        }
        return 0;
    }
    
    }
    signed main(){
        Miracle::main();
        return 0;
    }
    
    /*
       Author: *Miracle*
    */

转载于:https://www.cnblogs.com/Miracevin/p/10766319.html

[BJOI2019]: https://www.luogu.org/problemnew/show/P5320
[rqy]: https://www.luogu.org/blog/rqy/solution-p5320
[1399631-20190425090447726-16565619.png]: /images/20211223/c42f856fb3f54474b298e8c9bf1018a7.png
[1399631-20190425090519182-1233896492.png]: /images/20211223/8c713e8482c24b2ea3f7d92b78eb9ab5.png