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Author: MathProgrammer

2020/Div 3/617 Div 3/Explanations/Array with Odd Sum Explanation.txt

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+If there are an odd number of odd numbers, then the sum is odd
+
+If there are an even number of odd numbers and there is at least one even number,
+we can make the even number odd and get an odd sum
+
+If there are no odd numbers, the sum will always be even
+
+-----
+
+void solve()
+{
+    int no_of_elements;
+    cin >> no_of_elements;
+
+    int odd_count = 0;
+    for(int i = 1; i <= no_of_elements; i++)
+    {
+        int x;
+        cin >> x;
+
+        odd_count += (x%2 == 1);
+    }
+
+    cout << ((odd_count%2 == 1) || (odd_count > 0 && odd_count%2 == 0 && odd_count != no_of_elements) ? "YES\n" : "NO\n");
+}

2020/Div 3/617 Div 3/Explanations/Food Buying Explanation.txt

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+We can just be greedy and simulate the problem.
+
+We will use 10(X/10) coins. We will get (X/10) extra coins
+
+We will repeat the same with the (X/10) coins that we have now and keep doing this until we have at least 10 coins
+
+
+-----
+
+void solve()
+{
+    int budget;
+    cin >> budget;
+
+    long long no_of_moves = 0;
+    while(budget >= 10)
+    {
+        long long move = budget/10;
+
+        budget = budget - 10*move + move;
+
+        no_of_moves += 10*move;
+    }
+
+    no_of_moves += budget;
+    cout << no_of_moves << "\n";
+}

2020/Div 3/617 Div 3/Explanations/Yet Another Walking Robot Explanation.txt

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+We will keep track of (x, y) at every stage.
+
+Suppose we have done i steps and are at (x, y) and we were at (x, y) at step p too.
+
+Then, [p + 1, i] is a substring we can remove.
+
+For every i, we will check which was the last point at which we were at (x, y)
+
+Accordingly, we will check the size of the segment [L, R] and check if it is the smallest segment we have encountered so far
+
+-----
+
+void solve()
+{
+    int length;
+    string S;
+    cin >> length >> S;
+
+    map <pair <int, int>, int> position;
+    int minimum_distance = length + 1, left = 0, right = length + 1;
+
+    position[make_pair(0, 0)] = -1;
+    for(int x = 0, y = 0, i = 0; i < length; i++)
+    {
+        switch(S[i])
+        {
+            case 'L' : x++; break;
+            case 'R' : x--; break;
+            case 'U' : y++; break;
+            case 'D' : y--; break;
+        }
+
+        if(position.count(make_pair(x, y)) != 0)
+        {
+            int last_i = position[make_pair(x, y)] + 1;
+
+            int distance = (i) - (last_i - 1);
+
+            if(distance < minimum_distance)
+            {
+                minimum_distance = distance;
+                left = last_i ; right = i;
+            }
+        }
+
+        position[make_pair(x, y)] = i;
+    }
+
+    if(minimum_distance > length)
+    {
+        cout << "-1\n";
+        return;
+    }
+
+    cout << left + 1 << " " << right + 1 << "\n";
+}