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

*/

巴特西

BZOJ3329: Xorequ(二进制数位dp 矩阵快速幂)

题意

Sol

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