倍数除法算法c++题

Question

Problem Statement

You are given an array a of n integers.

You want to make all elements of a equal to zero by doing the following operation exactly three times:

Select a segment, for each number in this segment we can add a multiple of len to it, where len is the length of this segment (added integers can be different).

It can be proven that it is always possible to make all elements of a equal to zero.

Input The first line contains one integer n (1 ≤ n ≤ 100000): the number of elements of the array.

The second line contains n elements of an array a separated by spaces: a1 , a2 ,…, an (−10^9 ≤ ai ≤ 10^9).

Output The output should contain six lines representing three operations.

For each operation, print two lines:

The first line contains two integers l , r (1 ≤ l ≤ r ≤ n ): the bounds of the selected segment. The second line contains r − l +1 integers bl , bl +1,…, br (−10^18 ≤ bi ≤ 10^18): the numbers to add to al , al +1,…, ar , respectively; bi should be divisible by r − l +1.

Example

input

4

1 3 2 4

output

1 1

-1

3 4

4 2

2 4

-3 -6 -6

Code Solution

#include<bits/stdc++.h>
 
int64_t minv(int64_t a, int64_t m) {
    assert(0 < a && a < m);
    if (a == 1) return 1;
    return m - minv(m % a, a) * m / a;
}
 
int main() {
    using namespace std;
    ios_base::sync_with_stdio(false), cin.tie(nullptr);
 
    int N; cin >> N;
    vector<int64_t> A(N);
    for (int i = 0; i < N; i++) {
        cin >> A[i];
    }
 
    if (N == 1) {
        cout << 1 << ' ' << 1 << '\n' << -A[0] << '\n';
        cout << 1 << ' ' << 1 << '\n' << 0 << '\n';
        cout << 1 << ' ' << 1 << '\n' << 0 << '\n';
        exit(0);
    }
 
    int64_t invNm1 = minv(N-1, N);
    assert(invNm1 * (N-1) % N == 1);
 
    vector<int64_t> op1(N);
    vector<int64_t> op2(N-1);
    for (int i = 0; i < N-1; i++) {
        op2[i] = -A[i] % N * invNm1 % N;
        assert((A[i] + (op2[i] * (N-1))) % N == 0);
        op1[i] = -(A[i] + (op2[i] * (N-1))) / N;
        assert(A[i] + (op1[i] * N) + (op2[i] * (N-1)) == 0);
    }
 
    cout << 1 << ' ' << N << '\n';
    for (int i = 0; i < N; i++) {
        cout << op1[i] * N << " \n"[i+1==N];
    }
    cout << 1 << ' ' << N-1 << '\n';
    for (int i = 0; i < N-1; i++) {
        cout << op2[i] * (N-1) << " \n"[i+1==N-1];
    }
 
    cout << N << ' ' << N << '\n';
    cout << -A[N-1] << '\n';
 
    return 0;
}

Question

I believe the code is using an extended euclidean algorithm for minv, and I really need help to visualize what's going on with the minv part. Also, I'm having trouble following the logic inside the for loop with the op2 and op1 equations and how the code produces the output from the input provided. Would anyone please share explanations to the code with the equations and logic here?

Answer 1

minv function finds modular multiplicative inverse. Meaning, for a given number a , its inverse modulo N is a number b which satisfies a*b % N = 1 .

Extended Euclidian algorithm can be used to find such inverse.

As for:

cout << op1[i] * N << " \n"[i+1==N];

This is just index trickery with bools. "<space>\\n" has a type char[2] char[3] , thus indices can be used to select between a space and a newline. Bool can be used to index such array since bools are implicitly convertible to int s - true->1 , false->0 .