Question:
Levko loves tables that consist of n rows and n columns very much. He especially loves beautiful tables. A table is beautiful to Levko if the sum of elements in each row and column of the table equals k.
Unfortunately, he doesn't know any such table. Your task is to help him to find at least one of them.
Input:
The single line contains two integers, n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1000).
Output:
Print any beautiful table. Levko doesn't like too big numbers, so all elements of the table mustn't exceed 1000 in their absolute value.
If there are multiple suitable tables, you are allowed to print any of them.
Explanation:
Make a matrix of 'NxN' and make all the diagonals element of matrix as 'K' and rest all zero. automatically sum of rows and columns will be equal to sum of K.
Ex: N = 4 and K = 7
Answer / Output:
7 0 0 0
0 7 0 0
0 0 7 0
0 0 0 7
Code:
/* Deepak Yadav */
#pragma GCC optimize("O3,unroll-loops")
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace chrono;
using namespace __gnu_pbds;
#define fastio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define MOD 1000000007
#define MOD1 998244353
#define INF 1e18
#define nline "\n"
#define pb push_back
#define ppb pop_back
#define mp make_pair
#define ff first
#define ss second
#define PI 3.141592653589793238462
#define set_bits(x) __builtin_popcountll(x)
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define test long long T;cin>>T;while(T--)
//#define ll long long
#define fl(x, n) for (int x = 0; x < n; ++x)
#define cina(a, n) for(int i=0;i<n;i++){cin>>a[i];}
#ifdef Priyansh31dec
#define debug(x) cerr << #x<<" "; _print(x); cerr << endl;
#else
#define debug(x);
#endif
typedef long long ll;
typedef unsigned long long ull;
typedef long double lld;
typedef tree<pair<int, int>, null_type, less<pair<int, int>>, rb_tree_tag, tree_order_statistics_node_update > pbds; // find_by_order, order_of_key
void _print(ll t) {cerr << t;}
void _print(int t) {cerr << t;}
void _print(string t) {cerr << t;}
void _print(char t) {cerr << t;}
void _print(lld t) {cerr << t;}
void _print(double t) {cerr << t;}
void _print(ull t) {cerr << t;}
template <class T, class V> void _print(pair <T, V> p);
template <class T> void _print(vector <T> v);
template <class T> void _print(set <T> v);
template <class T, class V> void _print(map <T, V> v);
template <class T> void _print(multiset <T> v);
template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.ff); cerr << ","; _print(p.ss); cerr << "}";}
template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(set <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(multiset <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}
void _print(pbds v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
/*---------------------------------------------------------------------------------------------------------------------------*/
ll gcd(ll a, ll b) {if (b > a) {return gcd(b, a);} if (b == 0) {return a;} return gcd(b, a % b);}
ll expo(ll a, ll b, ll mod) {ll res = 1; while (b > 0) {if (b & 1)res = (res * a) % mod; a = (a * a) % mod; b = b >> 1;} return res;}
void extendgcd(ll a, ll b, ll*v) {if (b == 0) {v[0] = 1; v[1] = 0; v[2] = a; return ;} extendgcd(b, a % b, v); ll x = v[1]; v[1] = v[0] - v[1] * (a / b); v[0] = x; return;} //pass an arry of size1 3
ll mminv(ll a, ll b) {ll arr[3]; extendgcd(a, b, arr); return arr[0];} //for non prime b
ll mminvprime(ll a, ll b) {return expo(a, b - 2, b);}
bool revsort(ll a, ll b) {return a > b;}
ll combination(ll n, ll r, ll m, ll *fact, ll *ifact) {ll val1 = fact[n]; ll val2 = ifact[n - r]; ll val3 = ifact[r]; return (((val1 * val2) % m) * val3) % m;}
void google(int t) {cout << "Case #" << t << ": ";}
vector<ll> sieve(int n) {int*arr = new int[n + 1](); vector<ll> vect; for (int i = 2; i <= n; i++)if (arr[i] == 0) {vect.push_back(i); for (int j = 2 * i; j <= n; j += i)arr[j] = 1;} return vect;}
ll mod_add(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a + b) % m) + m) % m;}
ll mod_mul(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a * b) % m) + m) % m;}
ll mod_sub(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a - b) % m) + m) % m;}
ll mod_div(ll a, ll b, ll m) {a = a % m; b = b % m; return (mod_mul(a, mminvprime(b, m), m) + m) % m;} //only for prime m
ll phin(ll n) {ll number = n; if (n % 2 == 0) {number /= 2; while (n % 2 == 0) n /= 2;} for (ll i = 3; i <= sqrt(n); i += 2) {if (n % i == 0) {while (n % i == 0)n /= i; number = (number / i * (i - 1));}} if (n > 1)number = (number / n * (n - 1)) ; return number;} //O(sqrt(N))
ll getRandomNumber(ll l, ll r) {return uniform_int_distribution<ll>(l, r)(rng);}
/*--------------------------------------------------------------------------------------------------------------------------*/
bool Campare(const auto a, const auto b){
if(a.first==b.first) return a.second<b.second;
return a.first>b.first;
}
bool prime[1000001];
#define int long long int
int divisor(int n){
int cnt=0;
for(int i=1;i<=n;i++){
if(n%i==0) cnt++;
}
return cnt;
}
int compare5(pair<char,int> &p1, pair<char,int> &p2){
if(p1.second==p2.second)
return p1.second<p2.second;
return p1.first>p1.first;
}
void solve(){
int n,k; cin>>n>>k;
int arr[n][n] = {0};
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
arr[i][j] =0;
for(int i=0;i<n;i++)
arr[i][i] = k;
for(int i=0;i<n;i++){
for(int j=0;j<n;j++)
cout<<arr[i][j]<<" ";
cout<<endl;
}
}
signed main() {
#ifdef deepakyadav123
freopen("Error.txt", "w", stderr);
#endif
fastio();
auto start1 = high_resolution_clock::now();
//test
solve();
auto stop1 = high_resolution_clock::now();
auto duration = duration_cast<microseconds>(stop1 - start1);
#ifdef deepakyadav123
cerr << "Time: " << duration . count() / 1000 << endl;
#endif
}
Note: Read only Highlights of the code to understand directly rest are for understanding purpose only.
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