Ehab and the Expected GCD Problem
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<bits/stdc++.h> #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<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #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<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;} template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;} template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;} template<class T, class S> 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 ", ans); return 0; } /* */