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 <cstdio>
#include <cstring>
#include <cmath>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;
}

转载于:https://www.cnblogs.com/gccbuaa/p/6852180.html

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