[toj4111] Binomial efficient

求(nk)(mod232){n \choose k}\pmod {2^{32}}

2a1×3a2×5a3…2b1×3b2×5b3⋯×2c1×3c2×5c3…=2a1−b1−c1×3a2−b2−c2×5a3−b3−c3…\begin{align}
&{{2^{a1}\times3^{a2}\times5^{a3}\dots}\over{2^{b1}\times3^{b2}\times5^{b3}\dots\times2^{c1}\times3^{c2}\times5^{c3}\dots}}\\
&\\
&= 2^{a1-b1-c1}\times3^{a2-b2-c2}\times5^{a3-b3-c3}\dots\\
\end{align}

/* **********************************************

File Name: 4111.cpp

Auther: zhengdongjian@tju.edu.cn

Created Time: 2015年08月17日 星期一 19时46分13秒

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

typedef long long ll;
typedef unsigned int ui;
const ll MOD = 1LL << 32;
const int MAX = 1000007;
bool is_prime[MAX];
int prime[MAX], tot;
int times[MAX];

int get_prime() {
memset(is_prime, true, sizeof(is_prime));
for (int i = 2; i <= MAX / i; ++i) {
for (int j = i * i; j < MAX; j += i) {
is_prime[j] = false;
}
}

tot = 0;
for (int i = 2; i < MAX; ++i) {
if (is_prime[i]) {
prime[tot++] = i;
}
}
return tot;
}

int gao(int p, int n) {
//printf("gao %d, %d: ", p, n);
int sum = 0;
for (ll i = p; i <= n; i *= p) {
sum += n / i;
}
//printf("%d\n", sum);
return sum;
}

ui fast_pow(ui a, int n) {
ui res = 1;
while (n) {
if (n & 1) res *= a;
a *= a;
n >>= 1;
}
return res;
}

int main() {
ios::sync_with_stdio(false);
get_prime();
int T;
cin >> T;
while (T--) {
int n, k;
cin >> n >> k;
if (k == 0) {
cout << 1 << endl;
continue;
}
memset(times, 0, sizeof(times));
for (int i = 0; prime[i] <= n; ++i) {
times[i] += gao(prime[i], n);
times[i] -= gao(prime[i], k);
times[i] -= gao(prime[i], n - k);
}
ui res = 1;
for (int i = 0; prime[i] <= n; ++i) {
//printf("pair %d^%d\n", prime[i], times[i]);
res *= fast_pow(prime[i], times[i]);
res %= MOD;
}
cout << res << endl;
}
return 0;
}