题目
https://www.lydsy.com/JudgeOnline/status.php?user_id=7989
题意
n个数异或值为0的方案数,要求n个数为不大于m的质数。
思路
fwt 找n次后的异或和为0的情况
代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i,a,b) for(int i = (a);i <= (b);i++)
const int N = 100005;
const int mod = 1e9+7;
int rev=mod+1>>1;
ll A[N],B[N];
int p[N],top,vis[N];
void FWT(ll a[],int n)
{
for(int d=1; d<n; d<<=1)
{
for(int m=d<<1,i=0; i<n; i+=m)
{
for(int j=0; j<d; j++)
{
int x=a[i+j],y=a[i+j+d];
a[i+j]=(x+y)%mod,a[i+j+d]=(x-y+mod)%mod;
//xor:a[i+j]=x+y,a[i+j+d]=(x-y+mod)%mod;
//and:a[i+j]=x+y;
//or:a[i+j+d]=x+y;
}
}
}
}
void UFWT(ll a[],int n)
{
for(int d=1; d<n; d<<=1)
{
for(int m=d<<1,i=0; i<n; i+=m)
{
for(int j=0; j<d; j++)
{
int x=a[i+j],y=a[i+j+d];
a[i+j]=1LL*(x+y)*rev%mod,a[i+j+d]=(1LL*(x-y)*rev%mod+mod)%mod;
//xor:a[i+j]=(x+y)/2,a[i+j+d]=(x-y)/2;
//and:a[i+j]=x-y;
//or:a[i+j+d]=y-x;
}
}
}
}
void init(int n)
{
rep(i,2,n)
{
vis[i] = 1;
}
top = 0;
rep(i,2,n)
{
if(vis[i])
{
p[top++] = i;
}
for(int j = 0;j < top&&p[j]<=n/i;j++)
{
vis[i*p[j]] = 0;
if(i%p[j]==0) break;
}
}
}
ll qpow(ll x,ll n)
{
ll ans = 1;
while(n)
{
if(n%2) ans = ans*x%mod;
x = x*x%mod;
n /= 2;
}
return ans;
}
int main()
{
ll n,m;
init(100000);
while(scanf("%lld%lld",&n,&m) != EOF)
{
memset(A,0,sizeof(A));
int len = 1;
while(len<=m)
{
len<<=1;
}
for(int i = 0;p[i] <= m;i++)
{
A[p[i]] = 1;
}
FWT(A,len);
rep(i,0,len-1)
{
A[i] = qpow(A[i],n)%mod;
}
UFWT(A,len);
printf("%lld\n",(A[0]+mod)%mod);
}
return 0;
}