【BZOJ4589】Hard Nim(FWT)

题解:

由博弈论可以知道题目等价于求这$n$个数$\^$为0

快速幂$+fwt$

这样是$nlog^2$的 并不能过

而且得注意$m$的数组$\^$一下会生成$2m$

#include <bits/stdc++.h>
using namespace std;
#define rint register int
#define IL inline
#define rep(i,h,t) for(int i=h;i<=t;i++)
#define dep(i,t,h) for(int i=t;i>=h;i--)
#define ll long long
#define me(x) memset(x,0,sizeof(x))
#define mep(x,y) memcpy(x,y,sizeof(y))
#define mid (t<=0?(h+t-1)/2:(h+t)/2)
namespace IO{
    char ss[1<<24],*A=ss,*B=ss;
    IL char gc()
    {
        return A==B&&(B=(A=ss)+fread(ss,1,1<<24,stdin),A==B)?EOF:*A++;
    }
    template<class T> void read(T &x)
    {
        rint f=1,c; while (c=gc(),c<48||c>57) if (c=='-') f=-1; x=(c^48);
        while (c=gc(),c>47&&c<58) x=(x<<3)+(x<<1)+(c^48); x*=f; 
    }
    char sr[1<<24],z[20]; ll Z,C1=-1;
    template<class T>void wer(T x)
    {
        if (x<0) sr[++C1]='-',x=-x;
        while (z[++Z]=x%10+48,x/=10);
        while (sr[++C1]=z[Z],--Z);
    }
    IL void wer1()
    {
        sr[++C1]=' ';
    }
    IL void wer2()
    {
        sr[++C1]='\n';
    }
    template<class T>IL void maxa(T &x,T y) {if (x<y) x=y;}
    template<class T>IL void mina(T &x,T y) {if (x>y) x=y;} 
    template<class T>IL T MAX(T x,T y){return x>y?x:y;}
    template<class T>IL T MIN(T x,T y){return x<y?x:y;}
};
using namespace IO;
const int N=4e5+10;
const int M=5e4+10;
const int inv2=5e8+4;
const int mo=1e9+7;
int n;
void fwt_or(int *a,int o)
{
    for (int i=1;i<n;i*=2)
      for (int j=0;j<n;j+=(i*2))
        for (int k=0;k<i;k++)
          if (o==1) (a[i+j+k]+=a[j+k])%=mo;
          else (a[i+j+k]-=a[j+k])%=mo;
}
void fwt_and(int *a,int o)
{
    for (int i=1;i<n;i*=2)
      for (int j=0;j<n;j+=(i*2))
        for (int k=0;k<i;k++)
          if (o==1) (a[j+k]+=a[i+j+k])%=mo;
          else (a[j+k]-=a[i+j+k])%=mo;
}
void fwt_xor(int *a,int o)
{
    for (int i=1;i<n;i*=2)
      for (int j=0;j<n;j+=(i*2))
        for (int k=0;k<i;k++)
        {
            int x=a[j+k],y=a[i+j+k];
            a[j+k]=(x+y)%mo,a[i+j+k]=(x+mo-y)%mo;
            if (o==-1) a[j+k]=1ll*a[j+k]*inv2%mo,a[i+j+k]=1ll*a[i+j+k]*inv2%mo;
        }
}
int p[N],now[N],cnt,a[N],ans[N],b[N],m;
bool t[N];
void fwt(int *A,int *B)
{
    me(a); me(b);
    rep(i,0,m) a[i]=A[i],b[i]=B[i];
    fwt_xor(a,1); fwt_xor(b,1);
    rep(i,0,n) a[i]=1ll*a[i]*b[i]%mo;
    fwt_xor(a,-1);
    rep(i,0,m) B[i]=a[i];
}
/*void fwt(int *A,int *B)
{
    int C[N]={};
    rep(i,0,n)
      rep(j,0,n)
      {
          int k=1;
        C[i^j]=(C[i^j]+A[i]*B[j])%mo;
      }
    rep(i,0,n) B[i]=C[i];
}*/
int fsp(int x)
{
    mep(now,a); me(ans); ans[0]=1;
    while (x)
    {
        if (x&1) fwt(now,ans);
        fwt(now,now); x>>=1;
    }
    return ans[0];
}
int main()
{
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
    t[1]=1;
    rep(i,2,M)
    {
        if (!t[i]) p[++cnt]=i;
        for (int j=1;j<=cnt&&p[j]*i<=M;j++)
        {
          t[p[j]*i]=1;
          if (i%p[j]==0) break;
        }
    }
    int n1;
    while (cin>>n1>>m)
    {
        me(a);
        rep(i,1,cnt)
        {
          if (p[i]>m) break;
          a[p[i]]=1;
        }
        m*=2;
        for (n=1;n<=m;n<<=1);
        cout<<fsp(n1)<<endl;
    }
    return 0;
}

可以一次$fwt$之后最后再$ifwt$回去

复杂度$nlogn$

#include <bits/stdc++.h>
using namespace std;
#define rint register int
#define IL inline
#define rep(i,h,t) for(int i=h;i<=t;i++)
#define dep(i,t,h) for(int i=t;i>=h;i--)
#define ll long long
#define me(x) memset(x,0,sizeof(x))
#define mep(x,y) memcpy(x,y,sizeof(y))
#define mid (t<=0?(h+t-1)/2:(h+t)/2)
namespace IO{
    char ss[1<<24],*A=ss,*B=ss;
    IL char gc()
    {
        return A==B&&(B=(A=ss)+fread(ss,1,1<<24,stdin),A==B)?EOF:*A++;
    }
    template<class T> void read(T &x)
    {
        rint f=1,c; while (c=gc(),c<48||c>57) if (c=='-') f=-1; x=(c^48);
        while (c=gc(),c>47&&c<58) x=(x<<3)+(x<<1)+(c^48); x*=f; 
    }
    char sr[1<<24],z[20]; ll Z,C1=-1;
    template<class T>void wer(T x)
    {
        if (x<0) sr[++C1]='-',x=-x;
        while (z[++Z]=x%10+48,x/=10);
        while (sr[++C1]=z[Z],--Z);
    }
    IL void wer1()
    {
        sr[++C1]=' ';
    }
    IL void wer2()
    {
        sr[++C1]='\n';
    }
    template<class T>IL void maxa(T &x,T y) {if (x<y) x=y;}
    template<class T>IL void mina(T &x,T y) {if (x>y) x=y;} 
    template<class T>IL T MAX(T x,T y){return x>y?x:y;}
    template<class T>IL T MIN(T x,T y){return x<y?x:y;}
};
using namespace IO;
const int N=4e5+10;
const int M=5e4+10;
const int inv2=5e8+4;
const int mo=1e9+7;
int n;
void fwt_or(int *a,int o)
{
    for (int i=1;i<n;i*=2)
      for (int j=0;j<n;j+=(i*2))
        for (int k=0;k<i;k++)
          if (o==1) (a[i+j+k]+=a[j+k])%=mo;
          else (a[i+j+k]-=a[j+k])%=mo;
}
void fwt_and(int *a,int o)
{
    for (int i=1;i<n;i*=2)
      for (int j=0;j<n;j+=(i*2))
        for (int k=0;k<i;k++)
          if (o==1) (a[j+k]+=a[i+j+k])%=mo;
          else (a[j+k]-=a[i+j+k])%=mo;
}
void fwt_xor(int *a,int o)
{
    for (int i=1;i<n;i*=2)
      for (int j=0;j<n;j+=(i*2))
        for (int k=0;k<i;k++)
        {
            int x=a[j+k],y=a[i+j+k];
            a[j+k]=(x+y)%mo,a[i+j+k]=(x+mo-y)%mo;
            if (o==-1) a[j+k]=1ll*a[j+k]*inv2%mo,a[i+j+k]=1ll*a[i+j+k]*inv2%mo;
        }
}
int p[N],now[N],cnt,a[N],ans[N],b[N],m;
bool t[N];
void fwt(int *A,int *B)
{
    me(a); me(b);
    rep(i,0,m) a[i]=A[i],b[i]=B[i];
    fwt_xor(a,1); fwt_xor(b,1);
    rep(i,0,n) a[i]=1ll*a[i]*b[i]%mo;
    fwt_xor(a,-1);
    rep(i,0,m) B[i]=a[i];
}
/*void fwt(int *A,int *B)
{
    int C[N]={};
    rep(i,0,n)
      rep(j,0,n)
      {
          int k=1;
        C[i^j]=(C[i^j]+A[i]*B[j])%mo;
      }
    rep(i,0,n) B[i]=C[i];
}*/
int fsp(int x)
{
    mep(now,a); me(ans); ans[0]=1;
    fwt_xor(ans,1); fwt_xor(now,1);
    while (x)
    {
        if (x&1)
          rep(i,0,n) ans[i]=1ll*ans[i]*now[i]%mo;
        rep(i,0,n) now[i]=1ll*now[i]*now[i]%mo;
        x>>=1;
    }
    fwt_xor(ans,-1);
    return ans[0];
}
int main()
{
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
    t[1]=1;
    rep(i,2,M)
    {
        if (!t[i]) p[++cnt]=i;
        for (int j=1;j<=cnt&&p[j]*i<=M;j++)
        {
          t[p[j]*i]=1;
          if (i%p[j]==0) break;
        }
    }
    int n1;
    while (cin>>n1>>m)
    {
        me(a);
        rep(i,1,cnt)
        {
          if (p[i]>m) break;
          a[p[i]]=1;
        }
        m*=2;
        for (n=1;n<=m;n<<=1);
        cout<<fsp(n1)<<endl;
    }
    return 0;
}

 

posted @ 2018-12-14 14:36  尹吴潇  阅读(82)  评论(0编辑  收藏