HDU 4983 goffi and gcd (number theory)

HDU 4983 goffi and GCD

Idea: For the number topic, if K is 2 and N is 1, then only one type is possible. If other K> 2 is 0, you only need to consider k = 1, when k = 1, the factor n is enumerated, and then equal to the number of satisfied factors, then gcd (x, n) = The number of this factor is PHI (N/this factor), and then use the multiplication principle to calculate it.

Code:

#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;}

HDU 4983 goffi and gcd (number theory)