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解法

顯然可以二分答案

計算的時候用容斥原理即可

用莫比烏斯函數實現這個過程即可

代碼

#include <bits/stdc++.h>
#define LL long long
#define N 1000010
using namespace std;
int f[N], p[N], mu[N];
void sieve(){
    int len = 0, n = 1000000; mu[1] = 1;
    for (int i = 2; i <= n; i++) {
        if (!f[i]) p[++len] = i, mu[i] = -1;
        for (int j = 1; j <= len && i * p[j] <= n; j++) {
            f[i * p[j]] = true;
            if (i % p[j] == 0) {
                mu[i * p[j]] = 0;
                break;
            } else mu[i * p[j]] = -mu[i];
        }
    }
}
bool check(LL mid, LL k) {
    LL ret = 0;
    for (LL i = 1; i * i <= mid; i++)
        ret += mu[i] * (mid / (i * i));
    return mid - ret >= k;
}
int main() {
    sieve();
    LL k; cin >> k;
    LL l = 0, r = 210000000000ll, ans;
    while (l <= r) {
        LL mid = (l + r) >> 1;
        if (check(mid, k)) ans = mid, r = mid - 1;
            else l = mid + 1;
    }
    cout << ans << "\n";
    return 0;
}
 

bzoj 2986 Non-Squarefree Numbers 容斥原理+數學