Easy Math

推式子

$$\begin{gathered} \sum _{i=1}^m\mu (in) \\ \sum _{i=1}^m\mu (i\times \frac{n}{d}\times d)，d是n的一个质因子 \\ i,d互质项有(-\sum _{i=1}^m\mu (i\times \frac{n}{d}))，由于减去了多余的非互质项，所以加上， \\ -\sum _{i=1}^m\mu (i\times \frac{n}{d})+\sum _{i=1}^{\frac{m}{d}}\mu (i\times d\times \frac{n}{d}) \\ -\sum _{i=1}^m\mu (i\times \frac{n}{d})+\sum _{i=1}^{\frac{m}{d}}\mu (i\times n) \end{gathered}$$

由此我们可以递归加上杜教筛来得到答案：

边界条件$n==1$，$return{\sum }_{i=1}^m\mu (i)$。

$m==0,return0$。

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>using namespace std;typedef long long ll;
const int inf = 0x3f3f3f3f;const int N = 2e6 + 10;int prime[N], mu[N], cnt;bool st[N];void init() {mu[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {prime[cnt++] = i;mu[i] = -1;}for(int j = 0; j < cnt && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) break;mu[i * prime[j]] = -mu[i];}}for(int i = 1; i < N; i++) {mu[i] += mu[i - 1];}
}map<ll, ll> ans_s;ll S(ll n) {if(n < N) return mu[n];if(ans_s.count(n)) return ans_s[n];ll ans = 1;for(ll l = 2, r; l <= n; l = r + 1) {r = n / (n / l);ans -= (r - l + 1) * S(n / l);}return ans_s[n] = ans;
}ll F(ll n, ll m) {if(n == 1) return S(m);if(m == 0) return 0;for(int i = 0; 1ll * prime[i] * prime[i] <= n; i++) {if(n % prime[i] == 0) {return F(n, m / prime[i]) - F(n / prime[i], m);} }return F(n, m / n) - F(n / n, m);
}int main() {
//    freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);
//    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();ll n, m;scanf("%lld %lld", &m, &n);ll temp = n; for(int i = 0; i < cnt; i++) {int cnt = 0;while(n % prime[i] == 0) {n /= prime[i];cnt++;}if(cnt >= 2) {puts("0");return 0;}}printf("%lld\n", F(temp, m));return 0;
}