BZOJ 4589(Hard Nim-FWT测试)

题目：给n个不超过m的素数，求xor和=0的方案数，FWT变换裸题。
​​​题目​​

2关于F逆元的公式： inv(2)=(F+1)>>1
证：[(F+1)>>1]∗2(modF)=F+1(modF)=1

代码：

#include<bits/stdc++.h> 
using namespace std;
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#define
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma
#define
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll F=1000000007LL;
ll iv2=F+1>>1; //Õâ¸ö¼ÆËã˼·ºÃÆÀ 
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
void fwt(int*a,int n){
    int i,j,k,x;
    for(k=2;k<=n;k<<=1){
        for(i=0;i<n;i+=k){
            for(j=i;j<i+(k>>1);j++){
                x=a[j]+a[j+(k>>1)];if(x>=F) x-=F;
                a[j]=a[j]-a[j+(k>>1)];if(a[j]<0) a[j]+=F;
                a[j+(k>>1)]=x;
            }
        }
    }
}
void twf(int*a,int n){
    int i,j,k,x;
    for(k=n;k>=2;k>>=1){
        for(i=0;i<n;i+=k){
            for(j=i;j<i+(k>>1);j++){
                x=a[j]+a[j+(k>>1)];
                a[j+(k>>1)]=(int)(1LL*(a[j+(k>>1)]-a[j]+F)*iv2%F);
                a[j]=(int)(1LL*x*iv2%F);
            }
        }
    }
}
ll pow2(ll a,ll b){
    if (!b) return 1%F;
    ll p=pow2(a,b/2);
    p=mul(p,p);
    if (b&1) p=mul(p,a);
    return p;
}
#define
int b[MAXN],a[MAXN];
ll n,m;
int main()
{
//  freopen("bzoj4589.in","r",stdin);
//  freopen(".out","w",stdout);

    n=50000;
    MEM(b) b[1]=b[0]=1;
    Fork(i,2,n) if (!b[i]) {
        for(int j=2;i*j<=n;j++)
            b[i*j]=1;
    }   
    while(cin>>n>>m) {
        int len=1;
        while(len<=m) len<<=1;
        Rep(i,len) a[i]=(!b[i])&&(i<=m);
        fwt(a,len);
        Rep(i,len) a[i]=pow2(a[i],n);
        twf(a,len);
        printf("%d\n",a[0]);
    }


    return 0;
}