组合数取模就是求C(n,m)%p的值,当然根据n,m和p的取值范围不同,采取的方法也不一样。
1.P是素数:
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
typedef long long LL;
LL n,m,p;
LL quick_mod(LL a, LL b)
{
LL ans = 1;
a %= p;
while(b)
{
if(b & 1)
{
ans = ans * a % p;
b--;
}
b >>= 1;
a = a * a % p;
}
return ans;
}
LL C(LL n, LL m)
{
if(m > n) return 0;
LL ans = 1;
for(int i=1; i<=m; i++)
{
LL a = (n + i - m) % p;
LL b = i % p;
ans = ans * (a * quick_mod(b, p-2) % p) % p;
}
return ans;
}
LL Lucas(LL n, LL m)
{
if(m == 0) return 1;
return C(n % p, m % p) * Lucas(n / p, m / p) % p;
}
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
scanf("%I64d%I64d%I64d", &n, &m, &p);
printf("%I64d\n", Lucas(n,m));
}
return 0;
}
2.P不一定是素数也可能是合数
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
typedef long long LL;
const int N = 200005;
bool prime[N];
int p[N];
int cnt;
void isprime()
{
cnt = 0;
memset(prime,true,sizeof(prime));
for(int i=2; i<N; i++)
{
if(prime[i])
{
p[cnt++] = i;
for(int j=i+i; j<N; j+=i)
prime[j] = false;
}
}
}
LL quick_mod(LL a,LL b,LL m)
{
LL ans = 1;
a %= m;
while(b)
{
if(b & 1)
{
ans = ans * a % m;
b--;
}
b >>= 1;
a = a * a % m;
}
return ans;
}
LL Work(LL n,LL p)
{
LL ans = 0;
while(n)
{
ans += n / p;
n /= p;
}
return ans;
}
LL Solve(LL n,LL m,LL P)
{
LL ans = 1;
for(int i=0; i<cnt && p[i]<=n; i++)
{
LL x = Work(n, p[i]);
LL y = Work(n - m, p[i]);
LL z = Work(m, p[i]);
x -= (y + z);
ans *= quick_mod(p[i],x,P);
ans %= P;
}
return ans;
}
int main()
{
int T;
isprime();
cin>>T;
while(T--)
{
LL n,m,P;
cin>>n>>m>>P;
n += m - 2;
m--;
cout<<Solve(n,m,P)<<endl;
}
return 0;
}