精度被卡了，答案能算对，但是转成分数就会被卡精度/ll

#include <bits/stdc++.h>
#define re register
#define fst first
#define snd second
#define double long double

using namespace std;

typedef pair<int,int> pii;
const int N = 4e4 + 10;
double eps = 1e-10;
int n,k;
int f[N];

inline int read(){
    int r = 0,w = 1;
    char c = getchar();
    while (c < '0' || c > '9'){
        if (c == '-') w = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9'){
        r = (r << 3) + (r << 1) + (c ^ 48);
        c = getchar();
    }
    return r * w;
}

inline pii get(double x){
    int a = 0,b = 1;
    for (re int i = 1;i <= n;i++){
        int t = x * i;
        // cout << i << " " << t << " " << a << " " << b << " ???\n";
        if (fabs(1.0 * a / b - x) > fabs(1.0 * t / i - x)) a = t,b = i;
    }
    return {a,b};
}

inline bool check(double x){
    int num = 0;
    for (re int i = 1;i <= n;i++){
        f[i] = i * x;
        for (re int j = 2;j * j <= x;j++){
            if (i % j == 0){
                f[i] -= f[j];
                if (j != i / j) f[i] -= f[i / j];
            }
        }
        num += f[i];
    }
    return (num <= k);
}

int main(){
    n = read(),k = read();
    double l = 0.0,r = 1.0;
    while (r - l > eps){
        double mid = (l + r) / 2.0;
        if (check(mid)) l = mid;
        else r = mid;
    }
    // printf("%.3Lf\n",l);
    pii ans = get(l);
    printf("%d %d",ans.fst,ans.snd);
    return 0;
}