快速幂($m^n$)：
1 ll quickpow(ll m, ll n, ll k) 2 { 3 ll res = 1; 4 while (n > 0) 5 { 6 if (n & 1) 7 res = (res * m) % k; 8 m = (m * m) % k; 9 n = n >> 1; 10 } 11 return res; 12 }
快速乘($mn$)：
1 ll quickmul(ll m, ll n, ll k) 2 {
3 m = (m % k + k) % k; n = (n % k + k) % k; 4 ll res = 0; 5 while (n > 0) 6 { 7 if (n & 1) 8 res = (res + m) % k; 9 m = (m + m) % k; 10 n = n >> 1; 11 } 12 return res; 13 }
我觉得有必要换掉这个“龟速乘”
//不会爆long long
ll quickmul(ll a,ll b,ll mod) { ll ite = (1LL<<20)-1; return (a*(b>>20)%mod*(1ll<<20)%mod+a*(b&(ite))%mod)%mod; }
全部组合($C_i^j \% P,1leq ileq maxn,0leq jleq i$)
int C[maxn][maxn]; void comp(int n, int P) { memset(C, 0, sizeof(C)); for (int i = 0; i <= n; i++) { C[i][0] = 1; for (int j = 1; j <= i; j++) C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % P; } }
部分组合($C_n^0,C_n^1,cdots,C_n^n$)
void comp(int n) { C[0] = 1; for (int i = 1; i <= n; i++) C[i] = C[i - 1] * (n - i + 1) / i; //应该先乘后除，因为C[i-1]/i可能不是整数 }
单个组合($C_n^m$)
LL comp(LL n, LL m) { double ans = 1; for (int i = 0; i < m; i++) ans *= n - i; for (int i = 0; i < m; i++) ans /= i + 1; return (LL)(ans + 0.5); }