【板子】gcd、exgcd、乘法逆元、快速幂、快速乘、筛素数、快速求逆元、组合数

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

ll exgcd(ll l,ll r,ll &x,ll &y)
{
    if(r==0){x=1;y=0;return l;}
    else
    {
        ll d=exgcd(r,l%r,y,x);
        y-=l/r*x;
        return d;
    }
}

ll mod_inverse(ll a,ll m){
    ll x,y;
    if(exgcd(a,m,x,y)==1)//ax+my=1
        return (x%m+m)%m;
    return -1;//不存在
}

ll qpow(ll a,ll b,ll m){
    ll ans=1;
    ll k=a;
    while(b){
        if(b&1)ans=ans*k%m;
        k=k*k%m;
        b>>=1;
    }
    return ans;
}

ll qmul(ll a,ll b,ll m){
    ll ans=0;
    ll k=a;
    ll f=1;//f是用来存负号的
    if(k<0){f=-1;k=-k;}
    if(b<0){f*=-1;b=-b;}
    while(b){
        if(b&1)
            ans=(ans+k)%m;
        k=(k+k)%m;
        b>>=1;
    }
    return ans*f;
}

ll china(ll n, ll *a,ll *m) {
    ll M=1,y,x=0,d;
    for(ll i = 1; i <= n; i++) M *= m[i];
    for(ll i = 1; i <= n; i++) {
        ll w = M /m[i];
        exgcd(m[i], w, d, y);//m[i]*d+w*y=1
        x = (x + y*w*a[i]) % M;
    }
    return (x+M)%M;
}

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;
        }
    }
}

void inverse(){
    inv[1] = 1;
    for(int i=2;i<N;i++)
    {
        if(i >= M) break;
        inv[i] = (M-M/i)*inv[M%i]%M;
    }
}

ll fac[N]={1,1},inv[N]={1,1},f[N]={1,1};
ll C(ll a,ll b){
    if(b>a)return 0;
    return fac[a]*inv[b]%M*inv[a-b]%M;
}
void init(){//快速计算阶乘的逆元
    for(int i=2;i<N;i++){
        fac[i]=fac[i-1]*i%M;
        f[i]=(M-M/i)*f[M%i]%M;
        inv[i]=inv[i-1]*f[i]%M;
    }
}

ll C(ll n,ll m){
    if(m>n)return 0;
    ll ans=1;
    for(int i=1;i<=m;i++)
        ans=ans*(n-i+1)%M*qpow(i,M-2,M)%M;
    return ans;
}

#define N 100005
#define M 100007
ll n,m,fac[N]={1};
ll C(ll n,ll m){
    if(m>n)return 0;
    return fac[n]*qpow(fac[m],M-2,M)%M*qpow(fac[n-m],M-2,M)%M;//费马小定理求逆元
}
ll lucas(ll n,ll m){
    if(!m)return 1;
    return(C(n%M,m%M)*lucas(n/M,m/M))%M;
}
void init(){
    for(int i=1;i<=M;i++)
        fac[i]=fac[i-1]*i%M;
}