BZOJ3329: Xorequ(二进制数位dp 矩阵快速幂)

题意

题目链接

Sol

挺套路的一道题

首先把式子移一下项

\(x \oplus 2x = 3x\)

有一件显然的事情：\(a \oplus b \leqslant c\)

又因为\(a \oplus b + 2(a \& b) = c\)

那么\(x \& 2x = 0\)

也就是说，\(x\)的二进制表示下不能有相邻位

第一问直接数位dp即可

第二问比较interesting，设\(f[i]\)表示二进制为\(i\)的方案数，转移时考虑上一位选不选

如果能选，方案数为\(f[i - 2]\)

不选的方案数为\(f[i - 1]\)

#include<bits/stdc++.h>
#define LL long long
//#define int long long
#define file {freopen("a.in", "r", stdin); freopen("a.out", "w", stdout);}
using namespace std;
const int MAXN = 233, mod = 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;
}
LL N;
struct Matrix {
	int m[3][3];
	Matrix() {
		memset(m, 0, sizeof(m));
	}
	Matrix operator * (const Matrix &rhs) const {
		Matrix ans;
		for(int k = 1; k <= 2; k++)
			for(int i = 1; i <= 2; i++)
				for(int j = 1; j <= 2; j++)
					(ans.m[i][j] += 1ll * m[i][k] * rhs.m[k][j] % mod) %= mod;
		return ans;
	}
};
Matrix MatrixPow(Matrix a, LL p) {
	Matrix base;
	for(int i = 1; i <= 2; i++) base.m[i][i] = 1;
	while(p) {
		if(p & 1) base = base * a;
		a = a * a; p >>= 1;
	}
	return base;
}
LL num[MAXN], tot; LL f[MAXN][2];
LL dfs(int x, bool lim, bool pre) {
	if(!lim && (~f[x][pre])) return f[x][pre];
	if(x == 0) return 1;
	LL ans = 0;
	if(!pre && (num[x] == 1 || (!lim))) ans += dfs(x - 1, lim, 1);
	ans += dfs(x - 1, lim && num[x] == 0, 0);
	if(!lim) f[x][pre] = ans;
	return ans;
}
LL dp(LL x) {
	tot = 0;
	while(x) num[++tot] = x & 1, x >>= 1;
	return dfs(tot, 1, 0);
}
main() {
//	file;
	memset(f, -1, sizeof(f));
	int T = read();
	while(T--) {
		N = read();
		printf("%lld\n", dp(N) - 1);
		Matrix a;
		a.m[1][1] = 1; a.m[1][2] = 1;
		a.m[2][1] = 1; a.m[2][2] = 0;
		a = MatrixPow(a, N);
		printf("%d\n", (a.m[1][1] + a.m[1][2]) % mod);
	}
    return 0;
}
/*
1
5
*/