Detailed mathematical approach

Problem Summary

The sequence in the repository is generated by

$S_{n+1}\equiv S_n^2 \pmod{50515093},\qquad S_0=290797,$

and the working array starts at \(S_1=629527\). For the first \(N=2\cdot 10^9\) terms we must evaluate

$M(N)=\sum_{1\le l\le r\le N}\min(S_l,S_{l+1},\ldots,S_r).$

A direct scan over all subarrays is hopeless, and even the standard linear-time subarray-minimum technique cannot be run on \(2\cdot 10^9\) explicit values. The solution used in the C++, Python, and Java files therefore compresses the sequence to one period and then counts how often each position of that period owns a minimum.

Mathematical Approach

Step 1: Reduce the recurrence to one period

The map \(x\mapsto x^2 \bmod 50515093\) is deterministic on a finite state space, so once a value repeats, the future repeats as well. The implementation starts from \(S_1\), keeps generating values until \(S_1\) appears again, and obtains one full cycle

$a_1,a_2,\ldots,a_p,\qquad p=6308948.$

Hence the infinite sequence relevant to the problem is the periodic extension

$b_x=a_{((x-1)\bmod p)+1}\qquad (x\ge 1).$

Write the target length as

$N=qp+r,\qquad 0\le r<p.$

Then period index \(i\) appears

$\operatorname{occ}_i=q+\mathbf{1}_{i\le r}$

times inside the first \(N\) terms.

Step 2: Give every subarray minimum a unique owner

For a fixed absolute position \(x\), the classical contribution method counts how many subarrays have \(b_x\) as their representative minimum. Equal values require a tie-breaking rule. The code uses a strict comparison on the left and a non-strict comparison on the right, so ties are assigned to the rightmost occurrence of the minimum.

For a period index \(i\), with cyclic indexing modulo \(p\), define

$d_R(i)=\min\{k\ge 1: a_{i+k}\le a_i\},$

and define \(d_L(i)\) as the smallest \(k\ge 1\) with \(a_{i-k}<a_i\) if such a value exists within one full period to the left; otherwise set \(d_L(i)=0\).

The quantity \(d_R(i)\) is always at most \(p\), because the same value reappears one full period later. For \(d_L(i)\), looking back only one period is enough: if no strictly smaller value occurs there, periodicity implies that no strictly smaller value exists anywhere earlier.

Step 3: Build the span arrays in linear time

The implementations scan two consecutive copies of the period. A right-to-left monotone stack produces the next position with value \(\le a_i\), which gives \(d_R(i)\). A left-to-right monotone stack with the opposite strictness produces the previous position with value \(<a_i\), which gives \(d_L(i)\).

This is the usual subarray-minimum ownership trick, but adapted to a cyclic array. Two copies are sufficient because no relevant comparison can jump farther than one period.

Step 4: Contribution of one periodic index

The \(t\)-th occurrence of period index \(i\) in the prefix of length \(N\) sits at

$x_{i,t}=(t-1)p+i,\qquad 1\le t\le \operatorname{occ}_i.$

If \(L_{i,t}\) is the number of admissible left endpoints and \(R_{i,t}\) is the number of admissible right endpoints for subarrays owned by that occurrence, then its contribution is

$a_i\,L_{i,t}\,R_{i,t}.$

Every occurrence except the last one has the full right span \(d_R(i)\). Only the last occurrence may be cut by the end of the prefix. Its remaining tail length is

$\operatorname{tail}_i=\begin{cases} r-i+1, & i\le r,\\ p+r-i+1, & i>r, \end{cases}$

so the actual right factor on the last copy is

$R_i^{\text{last}}=\min(d_R(i),\operatorname{tail}_i).$

Step 5: Closed forms matching the code

If \(d_L(i)=0\), then no strictly smaller value exists anywhere to the left, so the left span is simply the absolute position itself:

$L_{i,t}=x_{i,t}=(t-1)p+i.$

When \(\operatorname{occ}_i=1\), this gives

$C_i=a_i\cdot i\cdot R_i^{\text{last}}.$

When \(\operatorname{occ}_i\ge 2\), the first \(\operatorname{occ}_i-1\) copies use the full right span \(d_R(i)\), so the code sums an arithmetic progression:

$\sum_{t=1}^{\operatorname{occ}_i-1}\big((t-1)p+i\big) =(\operatorname{occ}_i-1)i+p\frac{(\operatorname{occ}_i-2)(\operatorname{occ}_i-1)}{2}.$

Therefore

$C_i=a_i\left[ d_R(i)\left((\operatorname{occ}_i-1)i+p\frac{(\operatorname{occ}_i-2)(\operatorname{occ}_i-1)}{2}\right) +\big((\operatorname{occ}_i-1)p+i\big)R_i^{\text{last}} \right].$

If \(d_L(i)>0\), the first copy may be truncated on the left, but every later copy has the fixed left span \(d_L(i)\). Let

$L_i^{\text{first}}=\min(d_L(i),i).$

Then, for \(\operatorname{occ}_i=1\),

$C_i=a_i\,L_i^{\text{first}}\,R_i^{\text{last}},$

and for \(\operatorname{occ}_i\ge 2\),

$C_i=a_i\left[L_i^{\text{first}}d_R(i)+(\operatorname{occ}_i-2)d_L(i)d_R(i)+d_L(i)R_i^{\text{last}}\right].$

The required value is the sum over one period:

$\boxed{M(N)=\sum_{i=1}^{p} C_i.}$

How the Code Works

The C++ solution separates the work into build_period_sequence(), build_spans(), and sum_subarray_min_prefix(). The Python and Java versions mirror the same formulas and case split. The C++ file also checks the key intermediate facts used by the derivation:

$p=6308948,\qquad M(10)=432256955,\qquad M(10000)=3264567774119.$

Those checkpoints are important because they validate both the period construction and the strict/non-strict ownership convention before the final \(N=2\cdot 10^9\) computation is performed.

Complexity Analysis

Building one period takes \(O(p)\) time. Each monotone-stack pass is \(O(p)\), because every position is pushed and popped at most once. The final aggregation over all \(p\) period indices is again \(O(p)\). Thus the full algorithm runs in \(O(p)\) time and \(O(p)\) memory, with \(p=6308948\), while the huge input \(N\) only appears through the two integers \(q\) and \(r\).

Footnotes and References

1. Problem page: https://projecteuler.net/problem=375
2. Monotone stacks and subarray minimum ownership: cp-algorithms
3. Cycle detection in deterministic sequences: Wikipedia — Cycle detection
4. Modular arithmetic background: Wikipedia — Modular arithmetic