求 \({n+m \choose n}\bmod p\)
卢卡斯(Lucas)定理
\[{n \choose m}\equiv{n\bmod p \choose m\bmod p}\times{\lfloor{n\over p}\rfloor \choose \lfloor{m\over p}\rfloor}\pmod p \]
证明（感谢Lance1ot）
首先我们需要证明
\[{p\choose i}\equiv{p\over i}{p-1\choose i-1}\equiv 0\pmod p,(1\le i\le p-1) \]
由
\[{p\choose i}={p!\over i!(p-i)!}={p\over i} {(p-1)!\over(i-1)!(p-1-i+1)!} {p\over i} {(p-1)!\over(i-1)!(p-i)!}={p\over i}{p-1\choose i-1} \]
得证。
然后根据这种性质和二项式定理，我们马上得出
\[(1+x)^p\equiv {p\choose0}1^p+{p\choose1}x^{2}+....+{p\choose p}x^p\equiv {p\choose0}1^px^0+{p\choose p}1^0x^p\equiv 1+x^p\pmod p \]
然后我们接下来要求证
\[{a\choose b}\equiv {a_0\choose b_0}{a_1p\choose b_1p}{a_2p^2\choose b_2p^2}\dots\pmod p \]
令\(a=lp+r,b=sp+j\)
求证\({a\choose b}\equiv {l\choose s}{r\choose j}\pmod p\)然后利用性质递归求解就可以了。
继续从二次项定理出发
\[(1+x)^a=(1+x)^{lp} \cdot (1+x)^r \]
然后展开\((1+x)^{lp}\)
\[(1+x)^{lp} \equiv ((1+x)^p)^l \equiv (1+x^p)^l\pmod p \]
\[\therefore (1+x)^a \equiv (1+x^p)^l(1+x)^r\pmod p \]
观察项\(x^b\)的系数
\[\because {a^b\choose x^b}\equiv{l\choose s}x^{sp}{r\choose j}x^j\pmod p \]
\[\therefore {a\choose b}x^b \equiv {l\choose s}{r\choose j}x^b\pmod p \]
\[\therefore {a\choose b}\equiv {l\choose s}{r\choose j} \equiv {\lfloor {a\over p} \rfloor\choose \lfloor {b\over p} \rfloor}{a\bmod p\choose b\bmod p}\pmod p \]
得证
实现
#include <cstdio>
#define ll long long
#define re register
#define il inline
#define gc getchar
#define pc putchar
template <class T>
void read(T &x) {
re bool f = 0;
re char c = gc();
while ((c < '0' || c > '9') && c != '-') c = gc();
if (c == '-') f = 1, c = gc();
x = 0;
while (c >= '0' && c <= '9') x = x * 10 + (c ^ 48), c = gc();
f && (x = -x);
}
template <class T>
void print(T x) {
if (x < 0) pc('-'), x = -x;
if (x >= 10) print(x / 10);
pc((x % 10) ^ 48);
}
template <class T>
void prisp(T x) {
print(x);
pc(' ');
}
template <class T>
void priln(T x) {
print(x);
pc('\n');
}
ll fac[100005];
ll pow(ll b, int t, ll p) {
ll r;
for (r = 1; t; t >>= 1, b = (b * b) % p)
if (t & 1) r = (r * b) % p;
return r;
}
ll C(ll n, ll m, ll p) {
if (m > n) return 0;
return (fac[n] * pow(fac[m], p - 2, p) % p) * pow(fac[n - m], p - 2, p) % p;
}
ll lucas(ll n, ll m, ll p) {
if (m == 0) return 1;
return C(n % p, m % p, p) * lucas(n / p, m / p, p) % p;
}
int main() {
int t;
read(t);
while (t--) {
ll n, m, p;
read(n);
read(m);
read(p);
fac[0] = 1;
for (int i = 1; i <= p; ++i) fac[i] = (fac[i - 1] * i) % p;
priln(lucas(n + m, m, p));
}
}