伯努利数求幂和（板子）

Posted 2020-12-05 bibibi

ll qpow(ll a,ll b){
    ll res=1;
    while(b){
        if(b&1)
        res=res*a%mod;
        b>>=1;
        a=a*a%mod;
    }
    return res;
}
ll c[maxn][maxn], inv[maxn], B[maxn];
void exgcd(ll a, ll b, ll &x, ll &y){
    if(b == 0){
        x = 1;y = 0;
        return ;
    }
    ll x1, y1;
    exgcd(b, a%b, x1, y1);
    x = y1;
    y = x1 - (a/b)*y1;
}
void get_fac(){
    for(int i=0; i<maxn; i++){
        c[i][0] = 1; c[i][i] = 1;
    }
    for(int i=1; i<maxn; i++)
        for(int j=1; j<=i; j++)
            c[i][j] = (c[i-1][j]+c[i-1][j-1])%mod;
}
 
void get_inv(){
    for(int i=1; i<maxn; i++){
        ll x, y;
        exgcd(i, mod, x, y);
        x = (x%mod+mod)%mod;
        inv[i] = x;
    }
}
void get_bb(){
    B[0] = 1;
    for(int i=1; i<maxn-1; i++){
        ll tmp = 0;
        for(int j=0; j<i; j++)
            tmp = (tmp+c[i+1][j]*B[j])%mod;
        B[i] = tmp;
        B[i] = B[i]*(-inv[i+1]);
        B[i] = (B[i]%mod+mod)%mod;
    }
}
 
ll get(ll n,int k){//返回(1^k+2^k+...n^k)%mod
    ++n;
    n %= mod;
    ll ans = 0;
    for(int i=1; i<=k+1; i++){
        ans = (ans+((c[k+1][i]*B[k+1-i])%mod)*qpow(n,(ll)i))%mod;
        ans = (ans%mod+mod)%mod;
    }
    ans = ans*inv[k+1];
    ans = (ans%mod+mod)%mod;
    return ans;
}
void init(){
    get_fac();
    get_inv();
    get_bb();
}