【TopCoder 11469】—Nim（Fwt）

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$Nim$游戏先手必败就是所有堆异或和为$0$

选$k$个异或和为$0$就是$Fwt$点值$k$次方
把所有质数拿出来做一个$Fwt$就可以了

#include<bits/stdc++.h>
using namespace std;
#define gc getchar
inline int read(){
	char ch=gc();
	int res=0,f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
#define re register
#define pb push_back
#define cs const
#define pii pair<int,int>
#define fi first
#define se second
#define ll long long
cs int mod=1e9+7;
inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
inline void Add(int &a,int b){(a+=b)>=mod?(a-=mod):0;}
inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
inline void Dec(int &a,int b){(a-=b)<0?(a+=mod):0;}
inline int mul(int a,int b){return 1ll*a*b>=mod?1ll*a*b%mod:a*b;}
inline void Mul(int &a,int b){a=mul(a,b);}
inline int ksm(int a,int b,int res=1){
	for(;b;b>>=1,a=mul(a,a))(b&1)&&(res=mul(res,a));return res;
}
inline void chemx(ll &a,ll b){a<b?a=b:0;}
inline void chemn(int &a,int b){a>b?a=b:0;}
cs int N=120005;
int pr[N],tot;
bitset<N> vis;
inline void init(int n){
	for(int i=2;i<=n;i++){
		if(!vis[i])pr[++tot]=i;
		for(int j=1;j<=tot&&pr[j]*i<=n;j++){
			vis[i*pr[j]]=1;
			if(i%pr[j]==0)break;
		}
	}
}
inline void Fwt(int *f,int lim,int kd){
	for(int a0,a1,mid=1;mid<lim;mid<<=1){
		for(int i=0;i<lim;i+=(mid<<1))
		for(int j=0;j<mid;j++){
			a0=f[i+j],a1=f[i+j+mid];
			f[i+j]=add(a0,a1),f[i+j+mid]=dec(a0,a1);
		}
	}
	if(kd==-1)for(int i=0,inv=ksm(lim,mod-2);i<lim;i++)Mul(f[i],inv);
}
int k,l;
int a[N];
class Nim{
	public:
	inline int count(int K,int L){
		k=K,l=L;
		int lim=1;
		init(L);
		while(lim<=pr[tot])lim<<=1;
		for(int i=1;i<=tot;i++)a[pr[i]]=1;
		Fwt(a,lim,1);
		for(int i=0;i<lim;i++)a[i]=ksm(a[i],k);
		Fwt(a,lim,-1);
		return a[0];
	}
};