HDU 4983 Goffi and GCD（数论)

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
;
ll prime
, frc
, fn, cnt
;

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;
}