Programación Competitiva

#include <bits/stdc++.h>
using namespace std;

#define ar array
#define ll long long
 
const int MAX_N = 5005;
const ll MOD = 1e9 + 7;
const ll INF = 0x3f3f3f;

ll qexp(ll a, ll b, ll m) {
    ll res = 1;
    while (b) {
        if (b % 2) res = res * a % m;
        a = a * a % m;
        b /= 2;
    }
    return res;
}

vector<ll> fact, invf;

void precompute(int n) {
    fact.assign(n + 1, 1); 
    for (int i = 1; i <= n; i++) fact[i] = fact[i - 1] * i % MOD;
    invf.assign(n + 1, 1);
    invf[n] = qexp(fact[n], MOD - 2, MOD);
    for (int i = n - 1; i > 0; i--) invf[i] = invf[i + 1] * (i + 1) % MOD;
}

ll nCk(int n, int k) {
    if (k < 0 || k > n) return 0;
    return fact[n] * invf[k] % MOD * invf[n - k] % MOD;
    // return fact[n] * qexp(fact[k], MOD - 2, MOD) % MOD * qexp(fact[n - k], MOD - 2, MOD) % MOD;
}

// A trick to calculate large factorial without overflowing is to take log at every step when precompute and take exponential when calculating
// Don't need invf[] now because it is the same as negative log of fact
vector<double> log_fact;
void precompute_log(int n) {
    log_fact.assign(n + 1, 0.0);
    log_fact[0] = 0.0; 
    for (int i = 1; i <= n; i++) log_fact[i] = log_fact[i - 1] + log(i); 
}

ll log_nCk(int n, int k) { 
    if (k < 0 || k > n) return 0;
    return exp(log_fact[n] - log_fact[n - k] - log_fact[k]); 
}

int main(){
    int n,k;
    precompute(5005);
    while(scanf("%d %d",&n,&k)==2 && (n!=0 || k!=0)){
        for(int i=0; i<=k;i++){
            if(i==0) printf("%lld",nCk(n, i));
            else printf(" %lld",nCk(n, i));
        }
        printf("\n");
    }
    return 0;
}