另类求组合数

ll C[220][220];
void getC(){
    memset(C,0,sizeof(C));
    for(int i = 0; i < 220; i ++){
        C[i][0] = 1;
        for(int j = 1; j <= i; j ++){
            C[i][j] = (C[i-1][j-1]+C[i-1][j])%Mod;
        }
    }
}

 上面这种过时了，复杂度高。

下面这种快很多。

ll pow_mod(ll x, ll n){  
    ll res=1;  
    while(n>0){  
        if(n&1)res=res*x%mod;  
        x=x*x%mod;  
        n>>=1;  
    }  
    return res;  
}
ll C(int a, int b) {
    if(a < 0 || b < 0 || a < b) return 0;　　　
    return fac[a] * inv[b] % mod * inv[a-b] % mod;
}
fac[0] = 1;for(int i = 1; i <= mx; i ++) fac[i] = fac[i-1] * i %mod;
inv[mx] = pow_mod(fac[mx], mod-2); for(int i = mx-1; i >= 0; i --) inv[i] = inv[i+1] *(i+1) % mod;

 费马小定理：

const ll mod = 1e9+7;
const ll md = 1e9+6;
c[0][0] = 1;
for (int i = 1; i <= 1000; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++) {
        c[i][j] = (c[i-1][j-1] + c[i-1][j]) % md;
    }
}