题意

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
    
     #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;}/*15*/