Description

求∑i=1nikri${\displaystyle \sum _{i=1}^n{i}^k{r}^i}$

Solution

當r=1$r=1$時，就是個裸的自然數冪和問題。

如果r≠1$r\ne 1$，類似於求自然數冪和時的遞推做法，設S(k)=∑i=1nikri$S(k)=\sum _{i=1}^n{i}^k{r}^i$
那麼rS(k)=∑i=1nikri+1=∑i=1n+1(i−1)kri+1${\displaystyle rS(k)=\sum _{i=1}^n{i}^k{r}^{i+1}=\sum _{i=1}^{n+1}(i-1{)}^k{r}^{i+1}}$
所以有：

Code

inline LL Pow(LL a, LL b)
{
    LL Ans = 1;
    for (a %= Mod ; b; b >>= 1, (a *= a) %= Mod) if (b & 1
 
) (Ans *= a) %= Mod;
    return Ans;
}

inline LL C(int n, int m) { return fac[n] * ifac[m] % Mod * ifac[n - m] % Mod; }

LL S1(int n, int k)
{
    if (f[k]) return f[k];
    f[k] = pown[k + 1] - 1;
    rep(i, k) (f[k] -= C(k + 1, i) * S1(n, i) % Mod) %= Mod;
    ((((f[k] *= Pow(k + 1, Mod - 2)) %= Mod) += Mod) %= Mod);
    return
 
 f[k];
}

int main()
{
#ifdef hany01
    File("51nod1229");
#endif

    fac[0] = 1;
    For(i, 1, maxk - 5) fac[i] = fac[i - 1] * i % Mod;
    ifac[maxk - 5] = Pow(fac[maxk - 5], Mod - 2);
    Fordown(i, maxk - 5, 1) ifac[i - 1] = ifac[i] * i % Mod;

    for (static int T = read(); T --; ) {
        scanf("%lld", &n), k = read(), scanf("%lld", &r);

        if (r == 1) {
            pown[0] = 1;
            For(i, 1, k + 1) pown[i] = pown[i - 1] * (n + 1) % Mod;
            printf("%lld\n", S1(n, k));
            continue;
        }

        LL inv = Pow(r - 1, Mod - 2);
        S[0] = (Pow(r, n + 1) - r) % Mod * inv % Mod;
        For(i, 1, k) {
            S[i] = Pow(n, i) * Pow(r, n + 1) % Mod;
            For(j, 0, i - 1) (S[i] += (((i - j) & 1 ? -1 : 1) * C(i, j) * S[j] % Mod)) %= Mod;
            (S[i] *= inv) %= Mod;
        }
        printf("%lld\n", (S[k] + Mod) % Mod);
    }

    return 0;
}
//共看明月應垂淚，一夜鄉心五處同。
//    -- 白居易《望月有感》