BZOJ4589 Hard Nim（博弈+FWT）

　　即使n个数的异或为0。如果只有两堆，将质数筛出来设为1，做一个异或卷积即可。显然这个东西满足结合律，多堆时直接快速幂。可以在点值表示下进行。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
#define N (1<<17)
#define P 1000000007
#define inv2 500000004
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int T,n,m,f[N];
bool flag[N];
int ksm(int a,int k)
{
    int s=;
    for (;k;k>>=,a=1ll*a*a%P) if (k&) s=1ll*s*a%P;
    return s;
}
void FWT(int *a,int n,int op)
{
    for (int i=;i<=n;i<<=)
        for (int j=;j<n;j+=i)
            for (int k=j;k<j+(i>>);k++)
            {
                int x=a[k],y=a[k+(i>>)];
                a[k]=(x+y)%P,a[k+(i>>)]=(x-y+P)%P;
                if (op==) a[k]=1ll*a[k]*inv2%P,a[k+(i>>)]=1ll*a[k+(i>>)]*inv2%P;
            }
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4589.in","r",stdin);
    freopen("bzoj4589.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    flag[]=flag[]=;
    for (int i=;i<=;i++)
        for (int j=;j<=/i;j++)
        flag[i*j]=;
    while (scanf("%d %d",&n,&m)!=EOF)
    {
        int t=;while (t<=m) t<<=;
        for (int i=;i<=m;i++) f[i]=flag[i]^;
        for (int i=m+;i<t;i++) f[i]=;
        FWT(f,t,);
        for (int i=;i<t;i++) f[i]=ksm(f[i],n);
        FWT(f,t,);
        cout<<f[]<<endl;
    }
}