數論模板

組合數

const int MOD = 1e9 + 7;
const int M = 1e6 + 5;
int F[M], Finv[M], inv[M];//F是階乘，Finv是逆元的階乘 
void init()
{
    inv[1] = 1;
    for(int i = 2; i < M; i ++)
    {
        inv[i] = (MOD - MOD / i) * 1ll * inv[MOD % i] % MOD;
    }
    F[0] = Finv[0] = 1;
    for(int i = 1; i < M; i ++)
    {
        F[i] = F[i-1] * 1ll * i % MOD;
        Finv[i] = Finv[i-1] * 1ll * inv[i] % MOD;
    }
}
int C(int n, int m) // m <= n
{
    if(m < 0 || m > n) return 0;
    return F[n] * 1ll * Finv[n - m] % MOD * Finv[m] % MOD;
}

• 別忘了在主函數離調用打表函數啊QAQ

gcd

LL gcd(LL a, LL b)
{
    return b ? gcd(b, a%b) : a;
}

快速冪

LL pow_mod(LL a, LL b, LL p) //a的b次方求餘p 
{
    LL ret = 1;
    while(b)
    {
        if(b & 1) ret = (ret * a) % p;
        a = (a * a) % p;
        b >>= 1;
    }
    return ret;
}

快速乘

LL mul(LL a, LL b, LL p) //快速乘，計算a*b%p 
{
    LL ret = 0;
    while(b)
    {
        if(b & 1) ret = (ret + a) % p;
        a = (a + a) % p;
        b >>= 1;
    }
    return ret;
}

（不斷更新。。。感謝倉鼠學姐的博客orz）
http://www.cnblogs.com/linyujun/