首先我们能推出最优的开始一定是2 ^ n 或者 3 * 2 ^ n 的形式， 对于每一种形式我们将所有数字分类， 从后往前用组合数算出答案。

#include
     
      #define LL long long#define LD long double#define ull unsigned long long#define fi first#define se second#define mk make_pair#define PLL pair
      
       #define PLI pair
       
        #define PII pair
        
         #define SZ(x) ((int)x.size())#define ALL(x) (x).begin(), (x).end()#define fio ios::sync_with_stdio(false); cin.tie(0);using namespace std;const int N = 1e6 + 7;const int inf = 0x3f3f3f3f;const LL INF = 0x3f3f3f3f3f3f3f3f;const int mod = 1e9 + 7;const double eps = 1e-8;const double PI = acos(-1);template
         
           inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}template
          
            inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}template
           
             inline bool chkmax(T& a, S b) { return a < b ? a = b, true : false;}template
            
              inline bool chkmin(T& a, S b) { return a > b ? a = b, true : false;}int n;int a[N], tota;int b[N], totb;int F[N], Finv[N], inv[N];int ans;int cnt[21];inline int C(int n, int m) { if(n < m || m < 0) return 0; return 1LL * F[n] * Finv[m] % mod * Finv[n - m] % mod;}void calc(int n, int m) { int ret = F[cnt[0]], tmp = 0, sum = cnt[0]; for(int i = 1; i < m; i++) { tmp = ret; ret = 1LL * ret * cnt[i] % mod; ret = 1LL * ret * C(sum + cnt[i] - 1, cnt[i] - 1) % mod * F[cnt[i] - 1] % mod; sum += cnt[i]; } add(ans, ret);}void init() { F[0] = Finv[0] = inv[1] = 1; for(int i = 2; i < N; i++) inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod; for(int i = 1; i < N; i++) F[i] = 1LL * F[i - 1] * i % mod; for(int i = 1; i < N; i++) Finv[i] = 1LL * Finv[i - 1] * inv[i] % mod;}int main() { init(); a[++tota] = 2; for(int i = 2; ;i++) { a[i] = a[i - 1] * 2; tota++; if(a[i] > 1000000) break; } b[++totb] = 3; for(int i = 2; ;i++) { b[i] = b[i - 1] * 2; totb++; if(b[i] > 1000000) break; } scanf("%d", &n); int pa = upper_bound(a + 1, a + 1 + tota, n) - a - 1; int pb = upper_bound(b + 1, b + 1 + totb, n) - b - 1; int num = 0; for(int i = pa - 1; i >= 0; i--) { if(i) cnt[i] = n / (1 << i) - n / (1 << (i + 1)); else cnt[i] = n - n / 2; } calc(n, pa); if(pb == pa) { for(int o = pb - 1; o >= 0; o--) { int bit = pb - 1; for(int i = pb - 1; i >= 0; i--) { if(i > o) { cnt[i] = n / 3 / (1 << (bit - 1)) - n / 3 / (1 << bit); bit--; } else if(i < o) { if(i) { cnt[i] = n / (1 << (bit - 1)) - n / (1 << bit); bit--; } else cnt[i] = n - n / 2; } else { if(i) { cnt[i] = n / (1 << bit) - n / (1 << bit) / 3; } else cnt[i] = n - n / 3; } } calc(n, pb); } } printf("%d\n", ans); return 0;}/**/