HDU 4983 Goffi and GCD
思路：数论题。假设k为2和n为1。那么仅仅可能1种。其它的k > 2就是0种，那么事实上仅仅要考虑k = 1的情况了。k = 1的时候，枚举n的因子，然后等于求该因子满足的个数，那么gcd(x, n) = 该因子的个数为phi(n / 该因子)，然后再利用乘法原理计算就可以
代码：
#include#include #include typedef long long ll;const ll MOD = 1000000007;const int N = 35333;ll n, k, pn, vis[N];ll prime[N], frc[N], fn, cnt[N];void getprime() { pn = 0; for (ll i = 2; i < N; i++) { if (vis[i]) continue; prime[pn++] = i; for (ll j = i * i; j < N; j += i) vis[j] = 1; }}void getfrc(ll n) { fn = 0; for (ll i = 0; i < pn && n >= prime[i]; i++) { if (n % prime[i] == 0) { frc[fn] = prime[i]; cnt[fn] = 0; while (n % prime[i] == 0) { cnt[fn]++; n /= prime[i]; } fn++; } } if (n != 1) { frc[fn] = n; cnt[fn++] = 1; }}ll ans = 0;ll phi(ll n) { ll m = (ll)sqrt(n * 1.0); ll ans = n; for (ll i = 2; i <= m; i++) { if (n % i == 0) { ans = ans / i * (i - 1); while (n % i == 0) n /= i; } } if (n > 1) ans = ans / n * (n - 1); return ans;}void dfs(ll u, ll sum) { if (u == fn) { ll r = n / sum; ans = (phi(n / sum) * phi(sum) % MOD + ans) % MOD; return; } for (ll i = 0; i <= cnt[u]; i++) { dfs(u + 1, sum); sum *= frc[u]; }}ll solve() { getfrc(n); ans = 0; dfs(0, 1); return ans;}int main() { getprime(); while (~scanf("%I64d%I64d", &n, &k)) { if (n == 1) printf("1\n"); else if (k == 2) printf("1\n"); else if (k > 2) printf("0\n"); else { printf("%I64d\n", solve()); } } return 0;}