641B - Little Artem and Matrix - CodeForces Solution

// JAY SHREE RAM!!
#include<bits/stdc++.h>
using namespace std;
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
//using namespace __gnu_pbds;
#define int              long long int
#define cy               cout << "YES" << endl
#define cn               cout << "NO" << endl
#define c1               cout << "-1" << endl
#define all(v)           (v).begin(), (v).end()
#define rall(v)          (v).rbegin(), (v).rend()
#define maxi             *max_element
#define mini             *min_element
#define pb               push_back
#define sz(x)            (int)((x).size())
#define ff               first
#define ss               second
#define MOD              1000000007
//#const int inf           1e18+1
#define vi               vector<int>
#define input(v)         for(auto &item : v) cin >> item;
#define mod_add(a, b, m) (((a + b) % m) + m) % m
#define mod_mul(a, b, m) (((a * b) % m) + m) % m
#define mod_sub(a, b, m) (((a - b) % m) + m) % m
////#define ordered_set      tree<int, null_type, less<int>,
// ---------------------------------------------------------------------------------------
bool cmp(pair<int,int> a, pair<int,int> b) {
    if (a.first > b.first) {
        return true;
    } else if (a.first < b.first) {
        return false;
    } else {
        return a.second < b.second;
    }
}
bool isPrime(int n) {
    for(int i = 2; i * i <= n; i++) {
        if(n % i == 0) {
            return false;
        }
    }
    return true;
}

vi sieve(int n) {
    vi prime;
    vi vis(n + 1, 1);
    vis[0] = vis[1] = 0;
    for (int i = 2; i * i <= n; i++) {
        if (vis[i]) {
            for (int j = i * i; j <= n; j += i) {
                vis[j] = 0;
            }
        }
    }
    for (int i = 2; i <= n; i++) {
        if (vis[i]) {
            prime.pb(i);
        }
    }
    return prime;
}

int BinExp(int a, int b, int m) {
    int ans = 1;
    while (b > 0) {
        if (b & 1) {
            ans = (ans * 1LL * a) % m;
        }
        a = (a * 1LL * a) % m;
        b >>= 1;
    }
    return ans;
}

// ----------------------------------------------------------------------------------------

void solve() {
    // Your solution code goes here

    int n,m,q;
    cin>>n>>m>>q;

    vector<vector<int>> arr(n,vector<int> (m,0));

    /* for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            arr[i][j]=0;
        }
    }
     */

    class Operation
    {
        public:
        int type;
        int row,col;
        int val;

        Operation(int t,int r,int c,int v)
        {
            type=t;
            row=r;
            col=c;
            val=v;
        }
    };
    vector<Operation> oper;

    while(q--)
    {
        int type;
        cin>>type;

        if(type==1)
        {
        int r;cin>>r;
        oper.pb({type,r-1,0,0});
        
        }
        else if(type==2)
        {
            int c;cin>>c;
            oper.pb({type,0,c-1,0});
        }
        else
        {
            int r,c,x;
            cin>>r>>c>>x;
            oper.pb({type,r-1,c-1,x});

        }
    }
    // q-1 se bhi kar sakte.. but while loop ke baad q would exhaust soo..

    for(int i=sz(oper)-1;i>=0;i--)
    {
        Operation op=oper[i];

        if(op.type==1)
        {
            int temp=arr[op.row][m-1];
            // last ele ko store kiya
            for(int j=m-1;j>0;j--)
            {
                arr[op.row][j]=arr[op.row][j-1];
            }
            arr[op.row][0]=temp; // first banane ke liye
            // de circular shift xd
        }

        else if(op.type==2)
        {
            int temp=arr[n-1][op.col];
            for(int j=n-1;j>0;j--)
            {
                arr[j][op.col]=arr[j-1][op.col];
            }
            arr[0][op.col]=temp;
        }

        else
        {
            arr[op.row][op.col]=op.val;
        }
    }

    for(auto &it : arr)
    {
        for(auto & it1: it)
        {
            cout<<it1<<" ";

        }
        cout<<endl;
    }


}

int32_t main() {
    int t=1; //cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}