i want u to act as a n immpressive competetive programmer whos coedforces lgm and specializes in div2 contwsts, look at this div2a problem, dont fail like the rest, ill shou u the whole log of their failure
 want u to act as the most skilled competetive programmer, u take time to think and check if ur solution works and match up with the example tescase, u need to think thoroughly since u r in a contest.
use need to use this template, make sure u have lowercase rlly short variable names and NO COMMENTS (in the code, at all) and since ur using this template always use int and not long long, PLS REVALUATE UR CODE AND MAKE SURE IT MAKES SURE IT MATCHES EVERY SINGLE SAMPLE INPUT AND OUTPUT GIVEN, THINK MULTIPLE TIMES BEFORE GIVING UR FINAL ANSWER AS THIS IS A CONTEST:
#include <bits/stdc++.h>
#define int long long
using namespace std;

signed main()
{
cin.tie(0);
ios_base::sync_with_stdio(false);
}

think hard make sure it maches sample testcase, it's not hard, it needs focus, think from a more dp/math way!
C. Sakurako's Field Trip
time limit per test2 seconds
memory limit per test256 megabytes
Even in university, students need to relax. That is why Sakurakos teacher decided to go on a field trip. It is known that all of the students will be walking in one line. The student with index i
has some topic of interest which is described as ai
. As a teacher, you want to minimise the disturbance of the line of students.

The disturbance of the line is defined as the number of neighbouring people with the same topic of interest. In other words, disturbance is the number of indices j
(1≤j<n
) such that aj=aj+1
.

In order to do this, you can choose index i
(1≤i≤n
) and swap students at positions i
and n−i+1
. You can perform any number of swaps.

Your task is to determine the minimal amount of disturbance that you can achieve by doing the operation described above any number of times.

Input
The first line contains one integer t
(1≤t≤104
) — the number of test cases.

Each test case is described by two lines.

The first line contains one integer n
(2≤n≤105
) — the length of the line of students.
The second line contains n
integers ai
(1≤ai≤n
) — the topics of interest of students in line.
It is guaranteed that the sum of n
across all test cases does not exceed 2⋅105
.

Output
For each test case, output the minimal possible disturbance of the line that you can achieve.

Example
InputCopy
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3
OutputCopy
1
2
1
0
0
1
1
0
2
Note
In the first example, it is necessary to apply the operation to i=2
, thus the array will become [1,2,1,1,3]
, with the bold elements indicating those that have swapped places. The abstraction of this array is equal to 1
.

In the fourth example, it is sufficient to apply the operation to i=3
, thus the array will become [2,1,2,1,2,4]
. The abstraction of this array is equal to 0
.

In the eighth example, it is sufficient to apply the operation to i=3
, thus the array will become [1,4,1,5,3,1,3]
. The abstraction of this array is equal to 0
.



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

const int INF = 1e18;

signed main()
{
    cin.tie(0);
    ios_base::sync_with_stdio(false);
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        vector<int>a(n);
        for(auto &x:a) cin>>x;
        int m = n/2;
        vector<pair<int,int>>p;
        for(int i=0;i<m;i++) p.emplace_back(a[i],a[n-i-1]);
        vector<int>curr(n+1,INF);
        curr[0]=0;
        for(auto &[x,y]:p){
            vector<int>nxt(n+1,INF);
            for(int j=0;j<=n;j++) if(curr[j]<INF){
                // choice1: x then y
                int d = curr[j];
                if(j !=0 && j ==x) d+=1;
                if(x == y) d+=1;
                if(d < nxt[y]) nxt[y]=d;
                // choice2: y then x
                int dd = curr[j];
                if(j !=0 && j ==y) dd+=1;
                if(y == x) dd+=1;
                if(dd < nxt[x]) nxt[x]=dd;
            }
            curr = move(nxt);
        }
        int res=INF;
        if(n%2){
            int mid = a[m];
            for(int j=0;j<=n;j++) if(curr[j]<INF){
                int d = curr[j];
                if(j == mid) d+=1;
                res = min(res, d);
            }
        }
        else{
            for(int j=0;j<=n;j++) if(curr[j]<res) res = min(res, curr[j]);
        }
        cout<<res<<"\n";
    }
}

wa, THINK HARDER, NO COMMENTS, ITS NOT A DIFFICULT PROBLEM



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

signed main()
{
    cin.tie(0);
    ios_base::sync_with_stdio(false);
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        vector<int>a(n);
        for(auto &x:a) cin>>x;
        int m = n/2;
        vector<pair<int,int>>p;
        for(int i=0;i<m;i++) p.emplace_back(a[i],a[n-i-1]);
        int c0=0, c1=0;
        int l0=-1, l1=-1;
        for(auto &[x,y]:p){
            int nc0=1e18, nc1=1e18;
            // arrange as (x,y)
            int d0 = c0 + ((l0 == x) ? 1 : 0) + ((x == y) ? 1 : 0);
            nc0 = min(nc0, d0);
            d0 = c1 + ((l1 == x) ? 1 : 0) + ((x == y) ? 1 : 0);
            nc0 = min(nc0, d0);
            // arrange as (y,x)
            int d1 = c0 + ((l0 == y) ? 1 : 0) + ((y == x) ? 1 : 0);
            nc1 = min(nc1, d1);
            d1 = c1 + ((l1 == y) ? 1 : 0) + ((y == x) ? 1 : 0);
            nc1 = min(nc1, d1);
            c0 = nc0;
            c1 = nc1;
            l0 = y;
            l1 = x;
        }
        if(n%2){
            int mid = a[m];
            int res = 1e18;
            res = min(res, c0 + ((l0 == mid) ? 1 : 0));
            res = min(res, c1 + ((l1 == mid) ? 1 : 0));
            cout<<res<<"\n";
        }
        else{
            cout<<min(c0, c1)<<"\n";
        }
    }
}

input:
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3

this is the output ur giving
0
2
1
0
0
2
0
0
1

this is expected
1
2
1
0
0
1
1
0
2

focus, rework ur solution, think like lgm, and stop writing comments in the code. IT NEEDS TO BE LIKE EXPECTED OUTPUT



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

signed main()
{
    cin.tie(0);
    ios_base::sync_with_stdio(false);
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        vector<int>a(n);
        for(auto &x:a) cin>>x;
        int m = n/2;
        // Initialize previous states: two options, not swapped and swapped
        // Each state stores pair (last element, disturbances)
        // Initialize with no previous element
        pair<int,int> s0 = {-1,0};
        pair<int,int> s1 = {-1,0};
        for(int i=0;i<m;i++){
            int l = a[i];
            int r = a[n-i-1];
            // for each state, try both keep and swap
            vector<pair<int,int>> candidates;
            // from s0
            // keep
            int d = s0.second;
            if(s0.first != -1 && s0.first == l) d++;
            if(l == r) d++;
            candidates.emplace_back(r, d);
            // swap
            d = s0.second;
            if(s0.first != -1 && s0.first == r) d++;
            if(l == r) d++;
            candidates.emplace_back(l, d);
            // from s1
            // keep
            d = s1.second;
            if(s1.first != -1 && s1.first == l) d++;
            if(l == r) d++;
            candidates.emplace_back(r, d);
            // swap
            d = s1.second;
            if(s1.first != -1 && s1.first == r) d++;
            if(l == r) d++;
            candidates.emplace_back(l, d);
            // Find the minimum disturbance for each possible last element
            map<int,int> tmp;
            for(auto &[last, dist]: candidates){
                if(tmp.find(last)==tmp.end() || dist < tmp[last]) tmp[last]=dist;
            }
            // Update s0 and s1
            s0 = {-1, 1e18};
            s1 = {-1,1e18};
            for(auto &[last, dist]: tmp){
                if(s0.second > dist){
                    s0 = {last, dist};
                }
                else if(s1.second > dist){
                    s1 = {last, dist};
                }
            }
        }
        int res = 1e18;
        if(n%2){
            int mid = a[m];
            // From s0
            int d = s0.second;
            if(s0.first == mid) d++;
            res = min(res, d);
            // From s1
            d = s1.second;
            if(s1.first == mid) d++;
            res = min(res, d);
        }
        else{
            res = min(s0.second, s1.second);
        }
        cout<<res<<"\n";
    }
}

still wrong bro, u keep giving the wrong output, pls read again and look at ur mistakes and horror
C. Sakurako's Field Trip
time limit per test2 seconds
memory limit per test256 megabytes
Even in university, students need to relax. That is why Sakurakos teacher decided to go on a field trip. It is known that all of the students will be walking in one line. The student with index i
has some topic of interest which is described as ai
. As a teacher, you want to minimise the disturbance of the line of students.

The disturbance of the line is defined as the number of neighbouring people with the same topic of interest. In other words, disturbance is the number of indices j
(1≤j<n
) such that aj=aj+1
.

In order to do this, you can choose index i
(1≤i≤n
) and swap students at positions i
and n−i+1
. You can perform any number of swaps.

Your task is to determine the minimal amount of disturbance that you can achieve by doing the operation described above any number of times.

Input
The first line contains one integer t
(1≤t≤104
) — the number of test cases.

Each test case is described by two lines.

The first line contains one integer n
(2≤n≤105
) — the length of the line of students.
The second line contains n
integers ai
(1≤ai≤n
) — the topics of interest of students in line.
It is guaranteed that the sum of n
across all test cases does not exceed 2⋅105
.

Output
For each test case, output the minimal possible disturbance of the line that you can achieve.

Example
InputCopy
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3
OutputCopy
1
2
1
0
0
1
1
0
2
Note
In the first example, it is necessary to apply the operation to i=2
, thus the array will become [1,2,1,1,3]
, with the bold elements indicating those that have swapped places. The abstraction of this array is equal to 1
.

In the fourth example, it is sufficient to apply the operation to i=3
, thus the array will become [2,1,2,1,2,4]
. The abstraction of this array is equal to 0
.

In the eighth example, it is sufficient to apply the operation to i=3
, thus the array will become [1,4,1,5,3,1,3]
. The abstraction of this array is equal to 0
.



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

signed main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        vector<int>a(n);
        for(auto &x:a) cin>>x;
        int m = n/2;
        pair<int,int> s0 = {-1,0};
        pair<int,int> s1 = {-1,0};
        for(int i=0;i<m;i++){
            int x = a[i];
            int y = a[n-i-1];
            int nc0=1e18, nc1=1e18;
            int d0 = s0.second + ((s0.first == x) ? 1 : 0) + ((x == y) ? 1 : 0);
            nc0 = min(nc0, d0);
            int d1 = s1.second + ((s1.first == x) ? 1 : 0) + ((x == y) ? 1 : 0);
            nc0 = min(nc0, d1);
            int dd0 = s0.second + ((s0.first == y) ? 1 : 0) + ((y == x) ? 1 : 0);
            nc1 = min(nc1, dd0);
            int dd1 = s1.second + ((s1.first == y) ? 1 : 0) + ((y == x) ? 1 : 0);
            nc1 = min(nc1, dd1);
            s0 = {y, nc0};
            s1 = {x, nc1};
        }
        int res=1e18;
        if(n%2){
            int mid = a[m];
            res = min(s0.second + ((s0.first == mid) ? 1 : 0), s1.second + ((s1.first == mid) ? 1 : 0));
        }
        else{
            res = min(s0.second, s1.second);
        }
        cout<<res<<"\n";
    }
}

wrong still, use greedy approach man this is ks sad af



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

signed main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        vector<int>a(n);
        for(auto &x:a) cin>>x;
        int m = n/2;
        // Initialize two options: choose first arrangement or second
        // Each stores pair (last element, disturbance)
        pair<int,int> opt1 = { -1, 0 };
        pair<int,int> opt2 = { -1, 0 };
        for(int i=0;i<m;i++){
            int x = a[i];
            int y = a[n-i-1];
            // Current pair can be (x,y) or (y,x)
            // For each previous option, try both arrangements
            // and keep the minimum disturbances
            // Option (x,y)
            // From opt1
            int d1a = opt1.second + ((opt1.first != -1 && opt1.first == x) ? 1 : 0) + ((x == y) ? 1 : 0);
            // From opt2
            int d1b = opt2.second + ((opt2.first != -1 && opt2.first == x) ? 1 : 0) + ((x == y) ? 1 : 0);
            // Option (y,x)
            int d2a = opt1.second + ((opt1.first != -1 && opt1.first == y) ? 1 : 0) + ((y == x) ? 1 : 0);
            int d2b = opt2.second + ((opt2.first != -1 && opt2.first == y) ? 1 : 0) + ((y == x) ? 1 : 0);
            // Choose min for each arrangement
            pair<int,int> new_opt1 = {y, min(d1a, d1b)};
            pair<int,int> new_opt2 = {x, min(d2a, d2b)};
            opt1 = new_opt1;
            opt2 = new_opt2;
        }
        int res = LLONG_MAX;
        if(n%2){
            int mid = a[m];
            // From opt1
            int d1 = opt1.second + ((opt1.first == mid) ? 1 : 0);
            // From opt2
            int d2 = opt2.second + ((opt2.first == mid) ? 1 : 0);
            res = min(d1, d2);
        }
        else{
            res = min(opt1.second, opt2.second);
        }
        cout<<res<<"\n";
    }
}

STILL WRONG,


after looking at all these failures, talk to me and tell me how u will get the right solution!