正题

题目链接:https://darkbzoj.tk/problem/4589

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题目大意

求有多少个长度为$n$的数列满足它们都是不大于$m$的质数且异或和为$0$。

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解题思路

两个初始多项式$F[0]=1$，$G[prime\le m]=1$，然后答案就是$Fxor{G}^n$。然后$\text{FWT}$之后点值快速幂就好了。

时间复杂度$O(n\log n)$

$写水题有助于背板$

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code

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const ll N=(1<<16)+10,P=1e9+7,inv2=(P+1)/2;
ll n,k,m,f[N],g[N];
bool v[N];
void FWT(ll *f,ll op){if(op==-1)op=inv2;for(ll p=2;p<=n;p<<=1)for(ll k=0,len=p>>1;k<n;k+=p)for(ll i=k;i<k+len;i++){ll x=f[i],y=f[i+len];f[i]=(x+y)*op%P;f[i+len]=(x-y+P)*op%P;}return;
}
signed main()
{while(scanf("%lld%lld",&k,&m)!=EOF){memset(f,0,sizeof(f));memset(g,0,sizeof(g));memset(v,0,sizeof(v));n=1;while(n<=m)n<<=1;for(ll i=2;i<=m;i++){if(!v[i]){f[i]=1;for(ll j=i;j<=m;j+=i)v[j]=1;}}g[0]=1;FWT(g,1);FWT(f,1);while(k){if(k&1){for(ll i=0;i<n;i++)g[i]=g[i]*f[i]%P;}for(ll i=0;i<n;i++)f[i]=f[i]*f[i]%P;k>>=1;}FWT(g,-1);printf("%lld\n",g[0]);}return 0;
}