求组合数

第一类时mod为素数

 1 typedef long long LL;
 2 
 3 LL n,m,p;
 4 
 5 LL quick_mod(LL a, LL b)
 6 {
 7     LL ans = 1;
 8     a %= p;
 9     while(b)
10     {
11         if(b & 1)
12         {
13             ans = ans * a % p;
14             b--;
15         }
16         b >>= 1;
17         a = a * a % p;
18     }
19     return ans;
20 }
21 
22 LL C(LL n, LL m)
23 {
24     if(m > n) return 0;
25     LL ans = 1;
26     for(int i=1; i<=m; i++)
27     {
28         LL a = (n + i - m) % p;
29         LL b = i % p;
30         ans = ans * (a * quick_mod(b, p-2) % p) % p;
31     }
32     return ans;
33 }
34 
35 LL Lucas(LL n, LL m)
36 {
37     if(m == 0) return 1;
38     return C(n % p, m % p) * Lucas(n / p, m / p) % p;
39 }

p == prime

第二类时mod可以为合数

 1 typedef long long LL;
 2 const int N = 200005;
 3 
 4 bool prime[N];
 5 int p[N];
 6 int cnt;
 7 
 8 void isprime()
 9 {
10     cnt = 0;
11     memset(prime,true,sizeof(prime));
12     for(int i=2; i<N; i++)
13     {
14         if(prime[i])
15         {
16             p[cnt++] = i;
17             for(int j=i+i; j<N; j+=i)
18                 prime[j] = false;
19         }
20     }
21 }
22 
23 LL quick_mod(LL a,LL b,LL m)
24 {
25     LL ans = 1;
26     a %= m;
27     while(b)
28     {
29         if(b & 1)
30         {
31             ans = ans * a % m;
32             b--;
33         }
34         b >>= 1;
35         a = a * a % m;
36     }
37     return ans;
38 }
39 
40 LL Work(LL n,LL p)
41 {
42     LL ans = 0;
43     while(n)
44     {
45         ans += n / p;
46         n /= p;
47     }
48     return ans;
49 }
50 
51 LL Solve(LL n,LL m,LL P)
52 {
53     LL ans = 1;
54     for(int i=0; i<cnt && p[i]<=n; i++)
55     {
56         LL x = Work(n, p[i]);
57         LL y = Work(n - m, p[i]);
58         LL z = Work(m, p[i]);
59         x -= (y + z);
60         ans *= quick_mod(p[i],x,P);
61         ans %= P;
62     }
63     return ans;
64 }
65 
66 int main()
67 {
68     int T;
69     isprime();
70     cin>>T;
71     while(T--)
72     {
73         LL n,m,P;
74         cin>>n>>m>>P;
75         n += m - 2;
76         m--;
77         cout<<Solve(n,m,P)<<endl;
78     }
79     return 0;
80 }

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