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for(int i=0; i<n; i++)

vector<int> v(n);

sort(arr, arr+n);

dp[i][j] = max(dp[i-1][j], dp[i][j-1]);

while(l < r)

if(graph[u][v] == 1)

ans = (ans * base) % mod;

queue<pair<int,int>> q;

return gcd(a, b);

cout << result << endl;

1

0

1

1

0

1

0

1

0

O(n²)

O(log n)

O(n log n)

O(1)

O(2ⁿ)

O(n³)

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