Bzoj 1002 fjoi2007 rotavirus recurrence + High Precision

The rotavirus base defines how many n rotavirus viruses are available

This recursion is really not ...... So I chose to create a table.

First run the following program

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define M 110using namespace std;struct abcd{int to,next;bool ban;}table[M<<2];int head[M],tot=1;int n,ans;void Add(int x,int y){table[++tot].to=y;table[tot].next=head[x];head[x]=tot;}int fa[M],v[M],q[M],r,h;bool BFS(){int i;r=h=0;memset(v,0,sizeof v);memset(fa,-1,sizeof fa);q[++r]=0;while(r!=h){int x=q[++h];for(i=head[x];i;i=table[i].next)if(!table[i].ban){if(table[i].to==fa[x])continue;if(v[table[i].to])return 0;fa[table[i].to]=x;v[table[i].to]=1;q[++r]=table[i].to;}}if(r<=n)return 0;return 1;}void DFS(int x){if(x+x>tot){if( BFS() )++ans;return ;}table[x<<1].ban=table[x<<1|1].ban=0;DFS(x+1);table[x<<1].ban=table[x<<1|1].ban=1;DFS(x+1);}int main(){int i;while(1){memset(head,0,sizeof head);tot=1;ans=0;cin>>n;for(i=1;i<=n;i++)Add(0,i),Add(i,0),Add(i,i%n+1),Add(i%n+1,i);DFS(1);cout<<ans<<endl;}}

Simple enough, violent enough

Then, create a table ~ 14. The answer is as follows:

1 5 16 45 121 320 841 2205 5776 15125 39601 103680 271441 710645
Odd Number
1 16 121 841 5776 39601 271441
Root number
1 4 11 29 76 199 521
A [I] = A [I-1] * 3-A [I-2]
Even number
5 45 320 2205 15125 103680 710645
Divide by 5
1 9 64 441 3025 20736 142129
Root number
1 3 8 21 55 144 377
A [I] = A [I-1] * 3-A [I-2]

Then we can implement high-precision recursion.

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;struct abcd{    int x[100],cnt;    int& operator [] (int y)    {        return x[y];    }    void operator = (int y)    {        x[1]=y;        cnt=1;    }}f[100];abcd operator - (abcd x,abcd &y){    int i;    abcd z=f[0];    z.cnt=max(x.cnt,y.cnt);    for(i=1;i<=z.cnt;i++)    {        z[i]+=x[i]-y[i];        if(z[i]<0)            z[i+1]--,z[i]+=10;    }    while(z.cnt&&!z[z.cnt])        z.cnt--;    return z;}abcd operator * (abcd &x,abcd &y){    int i,j;    abcd z=f[0];    for(i=1;i<=x.cnt;i++)        for(j=1;j<=y.cnt;j++)            z[i+j-1]+=x[i]*y[j],z[i+j]+=z[i+j-1]/10,z[i+j-1]%=10;    z.cnt=x.cnt+y.cnt;    if(!z[z.cnt])        --z.cnt;    return z;}abcd operator * (abcd x,int y){    int i;    abcd z=f[0];    for(i=1;i<=x.cnt;i++)        z[i]+=x[i]*y,z[i+1]+=z[i]/10,z[i]%=10;    z.cnt=x.cnt;    if(z[z.cnt+1])        ++z.cnt;    return z;}ostream& operator << (ostream &os,abcd x){    int i;    for(i=x.cnt;i;i--)        os<<x[i];    return os;}int n;int main(){    int i;    cin>>n;    f[1]=1;    f[2]=n&1?4:3;    for(i=3;i+i<=n+1;i++)        f[i]=f[i-1]*3-f[i-2];    cout<<f[n+1>>1]*f[n+1>>1]*(n&1?1:5)<<endl;}

Bzoj 1002 fjoi2007 rotavirus recurrence + High Precision