The following is a very well-known problem, yet I recently found a nice solution involving convolutions.↵
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You're given $K$ and $n$ and you must compute for all $k \in [0, K]$↵
↵
$$S_k = 1^k + 2^k + \dots + n^k$$↵
↵
Obviously, we want it $mod$ some $M$, but let's ignore it for now.↵
↵
There're many ways of solving this problem, but here I focus on a particular one:↵
↵
We start with some standard stuff, by expanding $\sum_{i=1}^n (i+1)^{k + 1}$. Through the Binomial Theorem we get:↵
↵
↵
$$\displaystyle\sum_{i=1}^n (i+1)^{k + 1} = \sum_{i=1}^n \sum_{j=0}^{k+1} \binom{k+1}{j} i^j = \sum_{j=0}^{k+1} \binom{k+1}{j} \sum_{i=1}^n i^j = \sum_{i=1}^n i^{k + 1} + \binom{k + 1}{k} S_{k}+ \sum_{j=0}^{k-1} \binom{k+1}{j} S_j$$.↵
↵
Then, we can actually cancel out the LHS via some telescopic sum:↵
↵
↵
$$\displaystyle\sum_{i=1}^n (i+1)^{k + 1} = \sum_{i=1}^n i^{k + 1} + \binom{k + 1}{k} S_{k}+ \sum_{j=0}^{k-1} \binom{k+1}{j} S_j$$↵
↵
$$\displaystyle\sum_{i=1}^n (i+1)^{k + 1} - \sum_{i=1}^n i^{k + 1} = \binom{k + 1}{k} S_{k}+ \sum_{j=0}^{k-1} \binom{k+1}{j} S_j$$↵
↵
$$\displaystyle (n+1)^{k + 1} - 1 = (k + 1) S_{k}+ \sum_{j=0}^{k-1} \binom{k+1}{j} S_j$$↵
↵
From there, we get a recurrence:↵
↵
$$\displaystyle S_{k} = \frac{1}{k + 1} \left[(n+1)^{k + 1} - 1 - \sum_{j=0}^{k-1} \binom{k+1}{j} S_j \right]$$↵
↵
Now, let's expand the binomial coefficients and we get:↵
↵
$$\displaystyle S_{k} = \frac{1}{k + 1} \left[(n+1)^{k + 1} - 1 - (k+1)! \cdot \sum_{j=0}^{k-1} \frac{1}{(k+1-j)!} \cdot \frac{1}{j!} S_j \right]$$↵
↵
Now, let's define for each $t$: $A_{t}$ as $\frac{1}{(t+1)!}$ and $B_{t}$ as $\frac{1}{t!} S_t$. Notice that now the inner sum in the recurrence it's almost a convolution:↵
↵
We have pairs $A_{i}, B_{j}$, and the sum of the products $A_i \cdot B_j$ is contributing to $S_{i+j}$. The main issue is that we can't do a single convolution between $A$ and $B$ in order to compute $S$, because we need $S$ to define $B$.↵
↵
The alternative is to do some Divide and Conquer (thanks to Errichto who taught me this trick 2 years ago or more):↵
↵
Let's say we want to compute $S_k$ for every $k \in [l, r]$. Let's say $p=\lfloor \frac{l+r}{2} \rfloor$. We'll have a recursive algorithm:↵
↵
The idea is to first compute $S_k$ for each $k \in [l, p]$ recursively. Then to compute $\mathrm{B}[]$ for those values of $k$ and to apply the convolution. Then to update $S_k$ for all $k \in [p+1, r]$. Lastly, solve recursively for the other half ($[p+1,r]$).↵
↵
Something like this, in a very Pythonic style:↵
↵
~~~~~↵
def solve(l, r):↵
if l == r:↵
# base case, add the parts not related to the convolution↵
# use proper modular arithmetics ↵
# ...↵
return↵
mid = (l + r) // 2↵
solve(l, mid)↵
convolve(A, S[l...mid])↵
update S[mid+1...r] with the results from the convolution's results↵
solve(mid + 1, r)↵
~~~~~↵
↵
The final answer is computed by calling `solve(0, K)`.↵
Given that the sequences being convolved are of size $\mathcal{O}(r - l)$, we can convolve them in $\mathcal{O}((r - l)\log (r - l))$ time via FFT, giving a total time complexity of $\mathcal{O}(K\log^2 K)$.↵
↵
As for modular arithmetics, notice we need the multiplicative inverse of all numbers $j \in [1, K+1]$, hence the modulo better be good (a big prime and also FFT-friendly).↵
↵
That's all, I hope you like it. There are other ways of computing $S$, some of them quite interesting as well. I recommend going through them too.↵
↵
↵
You're given $K$ and $n$ and you must compute for all $k \in [0, K]$↵
↵
$$S_k = 1^k + 2^k + \dots + n^k$$↵
↵
Obviously, we want it $mod$ some $M$, but let's ignore it for now.↵
↵
There're many ways of solving this problem, but here I focus on a particular one:↵
↵
We start with some standard stuff, by expanding $\sum_{i=1}^n (i+1)^{k + 1}$. Through the Binomial Theorem we get:↵
↵
↵
$$\displaystyle\sum_{i=1}^n (i+1)^{k + 1} = \sum_{i=1}^n \sum_{j=0}^{k+1} \binom{k+1}{j} i^j = \sum_{j=0}^{k+1} \binom{k+1}{j} \sum_{i=1}^n i^j = \sum_{i=1}^n i^{k + 1} + \binom{k + 1}{k} S_{k}+ \sum_{j=0}^{k-1} \binom{k+1}{j} S_j$$.↵
↵
Then, we can actually cancel out the LHS via some telescopic sum:↵
↵
↵
$$\displaystyle\sum_{i=1}^n (i+1)^{k + 1} = \sum_{i=1}^n i^{k + 1} + \binom{k + 1}{k} S_{k}+ \sum_{j=0}^{k-1} \binom{k+1}{j} S_j$$↵
↵
$$\displaystyle\sum_{i=1}^n (i+1)^{k + 1} - \sum_{i=1}^n i^{k + 1} = \binom{k + 1}{k} S_{k}+ \sum_{j=0}^{k-1} \binom{k+1}{j} S_j$$↵
↵
$$\displaystyle (n+1)^{k + 1} - 1 = (k + 1) S_{k}+ \sum_{j=0}^{k-1} \binom{k+1}{j} S_j$$↵
↵
From there, we get a recurrence:↵
↵
$$\displaystyle S_{k} = \frac{1}{k + 1} \left[(n+1)^{k + 1} - 1 - \sum_{j=0}^{k-1} \binom{k+1}{j} S_j \right]$$↵
↵
Now, let's expand the binomial coefficients and we get:↵
↵
$$\displaystyle S_{k} = \frac{1}{k + 1} \left[(n+1)^{k + 1} - 1 - (k+1)! \cdot \sum_{j=0}^{k-1} \frac{1}{(k+1-j)!} \cdot \frac{1}{j!} S_j \right]$$↵
↵
Now, let's define for each $t$: $A_{t}$ as $\frac{1}{(t+1)!}$ and $B_{t}$ as $\frac{1}{t!} S_t$. Notice that now the inner sum in the recurrence it's almost a convolution:↵
↵
We have pairs $A_{i}, B_{j}$, and the sum of the products $A_i \cdot B_j$ is contributing to $S_{i+j}$. The main issue is that we can't do a single convolution between $A$ and $B$ in order to compute $S$, because we need $S$ to define $B$.↵
↵
The alternative is to do some Divide and Conquer (thanks to Errichto who taught me this trick 2 years ago or more):↵
↵
Let's say we want to compute $S_k$ for every $k \in [l, r]$. Let's say $p=\lfloor \frac{l+r}{2} \rfloor$. We'll have a recursive algorithm:↵
↵
The idea is to first compute $S_k$ for each $k \in [l, p]$ recursively. Then to compute $\mathrm{B}[]$ for those values of $k$ and to apply the convolution. Then to update $S_k$ for all $k \in [p+1, r]$. Lastly, solve recursively for the other half ($[p+1,r]$).↵
↵
Something like this, in a very Pythonic style:↵
↵
~~~~~↵
def solve(l, r):↵
if l == r:↵
# base case, add the parts not related to the convolution↵
# use proper modular arithmetics ↵
# ...↵
return↵
mid = (l + r) // 2↵
solve(l, mid)↵
convolve(A, S[l...mid])↵
update S[mid+1...r] with the results from the convolution
solve(mid + 1, r)↵
~~~~~↵
↵
The final answer is computed by calling `solve(0, K)`.↵
Given that the sequences being convolved are of size $\mathcal{O}(r - l)$, we can convolve them in $\mathcal{O}((r - l)\log (r - l))$ time via FFT, giving a total time complexity of $\mathcal{O}(K\log^2 K)$.↵
↵
As for modular arithmetics, notice we need the multiplicative inverse of all numbers $j \in [1, K+1]$, hence the modulo better be good (a big prime and also FFT-friendly).↵
↵
That's all, I hope you like it. There are other ways of computing $S$, some of them quite interesting as well. I recommend going through them too.↵
↵
