拉格朗日插值 求等次幂和模版

这个就是拉格朗日插值简化的公式

$$f(n)=\sum_{i=0}^{k+1}(-1)^{k-i+1}f(i)\frac{\prod_{j=0}^{i-1}(n-j)\prod_{j=i+1}^{k+1}(n-j)}{i!(k+1-i)!}$$

const ll mod=1e9+7;
const int maxn=1e3+10;

ll pref[maxn],pre_inv[maxn],f[maxn];

ll add(ll a,ll b){
    if ((a+=b)>=mod) {
        return a%mod;
    }else return a;
}

ll sub(ll a,ll b){
    if ((a-=b)<0) {
        return (a+mod)%mod;
    }else return a;
}

ll mul(ll a,ll b){
    if((a*=b)>=mod)
        return a%mod;
    else return a;
}

ll qmod(ll a,ll b){
    ll ans=1;
    while (b) {
        if (b&1) {
            ans=mul(ans, a);
        }
        b/=2;
        a=mul(a, a);
    }
    return ans;
}

void init(int k){
    pref[0]=pre_inv[0]=1;
    for (int i=1; i<=k+1; ++i) {
        pref[i]=mul(pref[i-1],i);
    }
    pre_inv[k+1]=qmod(pref[k+1], mod-2);
    for (int i=k; i>=1; --i) {
        pre_inv[i]=mul(pre_inv[i+1],(i+1));
    }
}

ll Lgr_polysum(int n,int k){
    f[0]=0;
    for (int i=1; i<=k+1; ++i) {
        f[i]=add(f[i-1], qmod(i, k));
    }
    ll hpre[maxn],tpre[maxn];
    hpre[0]=n;
    for (int i=1;i<=k+1 ; ++i) {
        hpre[i]=mul(hpre[i-1], n-i);
    }
    tpre[k+1]=sub(n,(k+1));
    for (int i=k; i>=0; --i) {
        tpre[i]=mul(tpre[i+1], sub(n,i));
    }
    ll ans=0;ll tmp;
    for (int i=0; i<=k+1; ++i) {
        if ((k+1-i)&1) {
            tmp=mul(f[i], mul(pre_inv[i],pre_inv[k+1-i]));
            if (i!=k+1)
                tmp=mul(tmp, tpre[i+1]);
            if(i)
                tmp=mul(hpre[i-1], tmp);
            ans=sub(ans, tmp);
        }else{
            tmp=mul(f[i],mul(pre_inv[i],pre_inv[k+1-i] ) );
            if (i!=k+1) {
                tmp=mul(tmp, tpre[i+1]);
            }
            if(i)
                tmp=mul(hpre[i-1], tmp);
            ans=add(ans, tmp);
        }
    }
    return ans;
}