BZOJ 3329: Xorequ [数位DP 矩阵乘法]

3329: Xorequ

题意：\(\le n \le 10^18\)和\(\le 2^n\)中满足\(x\oplus 3x = 2x\)的解的个数，第二问模1e9+7

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\(x\oplus 2x = 3x\) 不就是 \(x\oplus (x<<1) = (x<<1)+x\) 吗
异或是不进位的二进制加法，那么，没有相邻的1
然后第一问数位DP就很好搞了
第二问，n个数中选i个不能相邻，\(\sum\limits \binom{n+1-i}{i}\)
太大了没法算了， DP一下试试
\(f[i][0/1]\)i个数第i个是0/1的方案数
\(f[i][0]=f[i-1][1]+f[i-1][0],\ f[i][1]=f[i-1][0]\)
矩乘就好了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int N=65;
const ll P=1e9+7;
inline ll read(){
	char c=getchar();ll x=0,f=1;
	while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
	while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
	return x*f;
}

int a[N], len;
ll n, f[N][2];
ll dfs(int d, int last, int sky) { //printf("dfs %d %d %d\n",d,last,sky);
	if(d==0) return 1;
	if(!sky && f[d][last]!=-1) return f[d][last];
	int lim = sky ? a[d] : 1;
	ll now=0;
	now += dfs(d-1, 0, sky && 0==lim);
	if(last!=1 && 1<=lim) now += dfs(d-1, 1, sky && 1==lim);
	return sky ? now : f[d][last]=now;
}

struct Meow{
	ll a[2][2];
	Meow(){memset(a,0,sizeof(a));}
	void ini(){a[0][0]=a[1][1]=1;}
	ll* operator [](int x) {return a[x];}
};
inline void mod(ll &x) {if(x>=P) x-=P;}
Meow operator *(Meow a, Meow b) {
	Meow c;
	for(int i=0; i<2; i++)
		for(int k=0; k<2; k++) if(a[i][k])
			for(int j=0; j<2; j++) if(b[k][j])
				mod(c[i][j] += a[i][k]*b[k][j]%P);
	return c;
}
Meow operator ^(Meow a, ll b) { 
	Meow ans; ans.ini();
	for(; b; b>>=1, a=a*a)
		if(b&1) ans=ans*a;
	return ans;
}
int main() {
	freopen("in","r",stdin);
	memset(f,-1,sizeof(f));
	int T=read();
	Meow ans, f, g;
	g[0][0]=g[0][1]=g[1][0]=1; f[0][0]=f[1][0]=1;
	while(T--) {
		n=read(); len=0; ll m=n-1;
		while(n) a[++len]=n&1, n>>=1;
		printf("%lld\n", dfs(len, 0, 1)-1);
		ans=(g^m)*f;
		printf("%lld\n", (ans[0][0]+ans[1][0])%P);
	}
}