Problem 326: Modulo Summations

Project Euler Problem 326 Solution

EulerSolve provides an optimized solution for Project Euler Problem 326, Modulo Summations, with C++, Python, Java, and a step-by-step mathematical explanation.

Problem Summary The sequence is defined by $$a_1=1,\qquad a_n\equiv \sum_{k=1}^{n-1}k\,a_k\pmod n\qquad(n\ge 2).$$ For given $N$ and $M$, we must count how many subarrays of $$a_1,a_2,\dots,a_N$$ have sum divisible by $M$. The final instance is $$f(10^{12},10^6).$$ A direct $O(N)$ pass is impossible, so the key is to expose the hidden periodic structure. Mathematical Approach 1) Prefix sums turn divisibility into equal residues. Define prefix sums modulo $M$: $$P_t\equiv \sum_{i=1}^{t}a_i\pmod M,\qquad P_0=0.$$ Then for any $0\le i \lt j\le N$, $$\sum_{k=i+1}^{j}a_k\equiv 0\pmod M \iff P_i\equiv P_j\pmod M.$$ So if residue $r$ appears $c_r$ times among $$P_0,P_1,\dots,P_N,$$ then it contributes $$\binom{c_r}{2}$$ valid subarrays. Therefore $$f(N,M)=\sum_{r=0}^{M-1}\binom{c_r}{2}.$$ 2) Introduce the weighted running sum. Let $$W_n=\sum_{k=1}^{n}k\,a_k.$$ Then the recurrence is simply $$a_n=W_{n-1}\bmod n,\qquad W_n=W_{n-1}+n\,a_n.$$ This is exactly the pair of variables tracked in the C++ code. 3) The first terms reveal a 6-step block pattern. The code checks the first ten values $$1,1,0,3,0,3,5,4,1,9,$$ and if we continue a little further, the sequence groups naturally into blocks of length 6: $$[1,1,0,3,0,3],$$ $$[5,4,1,9,1,6],$$ $$[9,7,2,15,2,9],\dots$$ This is not an accident....

Detailed mathematical approach

Problem Summary

The sequence is defined by

$$a_1=1,\qquad a_n\equiv \sum_{k=1}^{n-1}k\,a_k\pmod n\qquad(n\ge 2).$$

For given $N$ and $M$, we must count how many subarrays of

$$a_1,a_2,\dots,a_N$$

have sum divisible by $M$. The final instance is

$$f(10^{12},10^6).$$

A direct $O(N)$ pass is impossible, so the key is to expose the hidden periodic structure.

Mathematical Approach

1) Prefix sums turn divisibility into equal residues.

Define prefix sums modulo $M$:

$$P_t\equiv \sum_{i=1}^{t}a_i\pmod M,\qquad P_0=0.$$

Then for any $0\le i \lt j\le N$,

$$\sum_{k=i+1}^{j}a_k\equiv 0\pmod M \iff P_i\equiv P_j\pmod M.$$

So if residue $r$ appears $c_r$ times among

$$P_0,P_1,\dots,P_N,$$

then it contributes

$$\binom{c_r}{2}$$

valid subarrays. Therefore

$$f(N,M)=\sum_{r=0}^{M-1}\binom{c_r}{2}.$$

2) Introduce the weighted running sum.

Let

$$W_n=\sum_{k=1}^{n}k\,a_k.$$

Then the recurrence is simply

$$a_n=W_{n-1}\bmod n,\qquad W_n=W_{n-1}+n\,a_n.$$

This is exactly the pair of variables tracked in the C++ code.

3) The first terms reveal a 6-step block pattern.

The code checks the first ten values

$$1,1,0,3,0,3,5,4,1,9,$$

and if we continue a little further, the sequence groups naturally into blocks of length 6:

$$[1,1,0,3,0,3],$$

$$[5,4,1,9,1,6],$$

$$[9,7,2,15,2,9],\dots$$

This is not an accident. For every integer $m\ge 0$, the exact formulas are

$$a_{6m+1}=4m+1,\qquad a_{6m+2}=3m+1,\qquad a_{6m+3}=m,$$

$$a_{6m+4}=6m+3,\qquad a_{6m+5}=m,\qquad a_{6m+6}=3m+3.$$

These identities can be proved by induction from the definition of $W_n$, and they are the real reason the problem becomes tractable.

4) Why this implies a period of $6M$.

Each formula above is affine in $m$. Therefore, modulo $M$, replacing $m$ by $m+M$ leaves every value unchanged. So

$$a_{n+6M}\equiv a_n\pmod M.$$

That gives periodicity of the term sequence modulo $M$.

5) Why the prefix residues also repeat with period $6M$.

We still need the prefix sums $P_t$, not just the terms $a_t$. For one full period, the block sum is

$$a_{6m+1}+\cdots+a_{6m+6}=18m+8.$$

Summing over $m=0,1,\dots,M-1$ gives

$$\sum_{n=1}^{6M} a_n = \sum_{m=0}^{M-1}(18m+8)=M(9M-1),$$

which is divisible by $M$. Therefore one whole period changes the prefix sum by

$$0\pmod M,$$

and hence the prefix-residue sequence $(P_t)$ itself has period $6M$.

6) Histogram over one period plus a tail.

Write

$$N=q(6M)+t,\qquad 0\le t \lt 6M.$$

Let $h_r$ be the number of times residue $r$ appears in one full period of

$$P_0,P_1,\dots,P_{6M-1},$$

and let $u_r$ be the number of appearances in the tail

$$P_0,P_1,\dots,P_t.$$

Then the total count of residue $r$ among $P_0,\dots,P_N$ is

$$c_r=q\,h_r+u_r.$$

Once these counts are known, the answer is just

$$f(N,M)=\sum_{r=0}^{M-1}\binom{c_r}{2}.$$

7) Worked checkpoint.

The code checks

$$f(10,10)=4,$$

and also

$$f(10^4,10^3)=97158.$$

These are excellent sanity checks for both the sequence generator and the residue-histogram logic.

Algorithm

1) Simulate exactly one period of length $6M$.

2) While simulating, record how often each prefix residue appears in the full period.

3) Also record the residue histogram for the first $t=N\bmod 6M$ positions.

4) Combine the two histograms via $c_r=q\,h_r+u_r$.

5) Sum $\binom{c_r}{2}$ over all residues.

Complexity Analysis

Only one period is simulated, so the running time is

$$O(M)$$

up to the constant factor 6, and the memory usage is

$$O(M).$$

The huge value of $N$ never appears as a loop bound.

Checks And Final Result

The checkpoints are

$$a_1,\dots,a_{10}=1,1,0,3,0,3,5,4,1,9,$$

$$f(10,10)=4,\qquad f(10^4,10^3)=97158.$$

For the target input, the final answer is

$$f(10^{12},10^6)=1966666166408794329.$$

Problem 326 source code

C++

#include <algorithm>
#include <array>
#include <cstdint>
#include <iostream>
#include <string>
#include <vector>

namespace {

using u64 = std::uint64_t;
using u128 = unsigned __int128;

std::string to_string_u128(u128 value) {
    if (value == 0) {
        return "0";
    }
    std::string s;
    while (value > 0) {
        const int digit = static_cast<int>(value % 10);
        s.push_back(static_cast<char>('0' + digit));
        value /= 10;
    }
    std::reverse(s.begin(), s.end());
    return s;
}

u128 count_pairs(const u64 N, const int M) {
    const u64 period = 6ULL * static_cast<u64>(M);
    const u64 full_cycles = N / period;
    const u64 remainder = N % period;

    std::vector<u64> freq_in_period(static_cast<std::size_t>(M), 0ULL);
    std::vector<u64> freq_in_tail(static_cast<std::size_t>(M), 0ULL);

    u128 weighted_sum = 0;  // sum_{k=1}^{n} k * a_k
    int prefix_mod = 0;     // sum_{i=1}^{n} a_i (mod M)

    for (u64 n = 0; n < period; ++n) {
        ++freq_in_period[static_cast<std::size_t>(prefix_mod)];
        if (n <= remainder) {
            ++freq_in_tail[static_cast<std::size_t>(prefix_mod)];
        }

        const u64 idx = n + 1ULL;
        const u64 a = (idx == 1ULL) ? 1ULL : static_cast<u64>(weighted_sum % idx);
        weighted_sum += static_cast<u128>(idx) * static_cast<u128>(a);

        prefix_mod += static_cast<int>(a % static_cast<u64>(M));
        if (prefix_mod >= M) {
            prefix_mod %= M;
        }
    }

    u128 answer = 0;
    for (int r = 0; r < M; ++r) {
        const u128 count = static_cast<u128>(full_cycles) * static_cast<u128>(freq_in_period[static_cast<std::size_t>(r)]) +
                           static_cast<u128>(freq_in_tail[static_cast<std::size_t>(r)]);
        answer += count * (count - 1) / 2;
    }

    return answer;
}

bool run_checkpoints() {
    {
        // Statement: first 10 values of a_n.
        std::array<u64, 10> expected = {1, 1, 0, 3, 0, 3, 5, 4, 1, 9};
        std::array<u64, 10> got{};

        u128 weighted_sum = 0;
        for (u64 n = 1; n <= 10; ++n) {
            const u64 a = (n == 1ULL) ? 1ULL : static_cast<u64>(weighted_sum % n);
            got[static_cast<std::size_t>(n - 1)] = a;
            weighted_sum += static_cast<u128>(n) * static_cast<u128>(a);
        }

        if (got != expected) {
            std::cerr << "Checkpoint failed: first 10 sequence elements mismatch\n";
            return false;
        }
    }

    if (count_pairs(10ULL, 10) != static_cast<u128>(4ULL)) {
        std::cerr << "Checkpoint failed: f(10,10)\n";
        return false;
    }

    if (count_pairs(10000ULL, 1000) != static_cast<u128>(97158ULL)) {
        std::cerr << "Checkpoint failed: f(10^4,10^3)\n";
        return false;
    }

    return true;
}

}  // namespace

int main(int argc, char** argv) {
    bool skip_checkpoints = false;
    for (int i = 1; i < argc; ++i) {
        const std::string arg(argv[i]);
        if (arg == "--skip-checkpoints") {
            skip_checkpoints = true;
        } else {
            std::cerr << "Unknown argument: " << arg << '\n';
            return 1;
        }
    }

    if (!skip_checkpoints && !run_checkpoints()) {
        return 2;
    }

    const u128 answer = count_pairs(1000000000000ULL, 1000000);
    std::cout << to_string_u128(answer) << '\n';
    return 0;
}

Python

def count_pairs(n_limit=1000000000000, m_val=1000000):
    period = 6 * m_val
    full_cycles = n_limit // period
    remainder = n_limit % period
    
    freq_in_period = [0] * m_val
    freq_in_tail = [0] * m_val
    
    weighted_sum = 0
    prefix_mod = 0
    
    for n in range(period):
        freq_in_period[prefix_mod] += 1
        if n <= remainder:
            freq_in_tail[prefix_mod] += 1
            
        idx = n + 1
        a = 1 if idx == 1 else weighted_sum % idx
            
        weighted_sum += idx * a
        prefix_mod = (prefix_mod + a % m_val) % m_val
        
    answer = 0
    for r in range(m_val):
        count = full_cycles * freq_in_period[r] + freq_in_tail[r]
        answer += count * (count - 1) // 2
        
    return str(answer)

def solve():
    return count_pairs()

if __name__ == '__main__':
    print(solve())

Java

import java.util.*;

public class Euler326 {
    public static String solve() {
        long limitN = 1000000000000L;
        int mVal = 1000000;

        long period = 6L * mVal;
        long fullCycles = limitN / period;
        long remainder = limitN % period;

        long[] freqInPeriod = new long[mVal];
        long[] freqInTail = new long[mVal];

        long sumLow = 0;
        long sumHigh = 0;
        int prefixMod = 0;

        for (long n = 0; n < period; n++) {
            freqInPeriod[prefixMod]++;
            if (n <= remainder) {
                freqInTail[prefixMod]++;
            }

            long idx = n + 1;
            long a;
            if (idx == 1) {
                a = 1;
            } else {
                long r64 = (Long.remainderUnsigned(-1L, idx) + 1) % idx;
                long hMod = Long.remainderUnsigned(sumHigh, idx);
                long lMod = Long.remainderUnsigned(sumLow, idx);
                a = (hMod * r64 + lMod) % idx;
            }

            long val = idx * a;
            sumLow += val;
            if (Long.compareUnsigned(sumLow, val) < 0) {
                sumHigh++;
            }

            prefixMod = (int) ((prefixMod + a % mVal) % mVal);
        }

        // Java 128 bit answer... wait!
        // answer can be (6,000,000)^2 / 2 ~ 1.8 * 10^13 ? No!
        // max value of answer:
        // sum_r (count_r * count_r / 2)
        // count_r is full_cycles * freq_in_period + freq_in_tail
        // fullCycles = 10^12 / 6M = 166,666.
        // sum(freq) = 6,000,000.
        // So sum(count) = 10^12.
        // answer is at worst 10^12 * 10^12 / 2 = 5 * 10^{23}.
        // This EXCEEDS Long.MAX_VALUE ~ 9 * 10^{18}.
        // We MUST use BigInteger for the final answer accumulation!

        java.math.BigInteger answer = java.math.BigInteger.ZERO;

        java.math.BigInteger fcBig = java.math.BigInteger.valueOf(fullCycles);

        for (int r = 0; r < mVal; r++) {
            java.math.BigInteger count = fcBig.multiply(java.math.BigInteger.valueOf(freqInPeriod[r]))
                    .add(java.math.BigInteger.valueOf(freqInTail[r]));
            java.math.BigInteger terms = count.multiply(count.subtract(java.math.BigInteger.ONE))
                    .divide(java.math.BigInteger.valueOf(2));
            answer = answer.add(terms);
        }

        return answer.toString();
    }

    public static void main(String[] args) {
        System.out.println(solve());
    }
}