lucas: 计算 C 时 当 n, m范围大 mod范围小时可以将 n m 缩减至 mod 范围内计算
预处理: 当 n m 范围小 mod 范围大时可以预处理阶乘逆元 O(1)计算
const int FN = 1e5 + 10;
const int MOD = 1e9 + 7;
ll fac[FN] = {1, 1}, inv[FN] = {1,1}, f[FN] = {1,1};
ll C(ll a,ll b)
{
if(b > a) return 0;
return fac[a] * inv[b] % MOD * inv[a - b] % MOD;
}
void init() //预处理逆元和阶乘
{
for(int i = 2; i < FN; i++)
{
fac[i] = fac[i - 1] * i % MOD;
f[i] = (MOD - MOD / i) * f[MOD % i] % MOD;
inv[i] = inv[i - 1] * f[i] % MOD;
}
}
lucas
C m
n = lucas(a, b)
ll mulit(ll a, ll b, ll m)
{
ll ans = 0;
while(b)
{
if(b & 1)
ans = (ans + a) % m;
a = (a << 1) % m;
b >>= 1;
}
return ans;
}
ll quick_mod(ll a, ll b, ll m)
{
ll ans = 1;
while(b)
{
if(b&1)
ans = mulit(ans, a, m);
a = mulit(a, a, m);
b >>= 1;
}
return ans;
}
ll comp(ll a, ll b, ll m)
{
if(a < b)
return 0;
if(a == b)
return 1;
if(b > a - b)
b = a - b;
ll ans = 1, ca = 1, cb = 1;
for(int i = 0; i < b; i++)
{
ca = ca * (a - i) % m;
cb = cb * (b - i) % m;
}
ans = ca * quick_mod(cb, m - 2, m) % m;
return ans;
}
ll lucas(ll a,ll b,ll m)
{
ll ans = 1;
while(a && b)
{
ans = (ans * comp(a % m, b % m, m)) % m;
a /= m;
b /= m;
}
return ans;
}