THIS CODE SHOWS WA ON LAST TEST!

general
1 
#include <bits/stdc++.h>
using namespace std;
#define int long long

void solve() {
    int n;
    cin >> n;
    int a[n], b[n];
    for (int& i : a) cin >> i;
    for (int& i : b) cin >> i;

    bool can = true;
    for (int i = 0; i < n - 3; i++) {
        if (b[i] == a[i]) continue;
        //simulate the operation
        if ((b[i] - a[i]) == (a[i + 1] + a[i + 2])) {
            a[i] += a[i + 1] + a[i + 2];
            a[i + 3] += a[i + 1] + a[i + 2];
            int t = a[i + 1];
            a[i + 1] = -a[i + 2];
            a[i + 2] = -t;
        }
        else {
            can = false;
            break;
        }
    }

    for (int i = 0; i < n; i++) {
        can &= a[i] == b[i];
    }
    cout << (can ? "yes" : "no") << "\n";
}

signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    int t;
    cin >> t;
    while (t--) {
        solve();
    }
}

Problem Link: Operating on A Practice Coding Problem - CodeChef

Please Help. Thank You.

2

Well, it should.

Your code assumes that the operation should be made from left to right, and that wasn’t stated in the problem. This is a counter example:

A = [8, 1, -8, 4, 7]
B = [1, 4, 3, 1, 3]

According to your code:
8 != 1
and
8 - 1 != 1 - 8

So, supposedly, it’s not possible.
But that’s not true.

I can perform the operation in index 2 (where it’s the num “1”)
So A becomes:
A = [8, -3, -4, 8, 3]

Then, perform the operation in index 1
A becomes:
A = [1, 4, 3, 1, 3], same as B.

There seems to be a correlation between the “apply this operation” problems and prefix sums. Specially, how does the prefix sums behaves when a given operation is done?

I wrote this:

#include <bits/stdc++.h> 
#define ll long long 
using namespace std; 
 
void solve(){ 
 
    ll N;
    cin >> N;
    
    vector<ll> A(N);
    vector<ll> B(N);
    for(ll i=0; i<N; i++){
        cin >> A[i];
    }
    for(ll i=0; i<N; i++){
        cin >> B[i];
    }
    
    vector<ll> preA(1, A[0]);
    vector<ll> preB(1, B[0]);
    for(ll i=1; i<N; i++){
        preA.push_back(preA[i-1] + A[i]);
        preB.push_back(preB[i-1] + B[i]);
    }
    
    map<ll,ll> oddA;
    map<ll,ll> oddB;
    map<ll,ll> evenA;
    map<ll,ll> evenB;
    for(ll i=0; i<N-1; i++){
        if(i % 2 == 0){
            evenA[preA[i]]++;
            evenB[preB[i]]++;
        }
        else{
            oddA[preA[i]]++;
            oddB[preB[i]]++;
        }
    }
    
    if ((evenA == evenB) && (oddA == oddB) && (preA[N-1] == preB[N-1])){
        cout << "YES\n";
    }
    else{
        cout << "NO\n";
    }
    
} 
 
int main() { 
    ios_base::sync_with_stdio(false); 
    cin.tie(NULL); 
    cout.tie(NULL); 
    int T; 
 
    cin >> T; 
    while(T--){ 
        solve(); 
    } 
}