Merge pull request #324 from suzzzal/day3sol2

Day3sol2

Author: zonedoutdg

Problems/Mathematics/Day-02/sol/Sujal_Kshatri/day2sol1.cpp

@@ -0,0 +1,51 @@
+// submission link - https://codeforces.com/problemset/submission/1771/355587277
+
+#include <bits/stdc++.h>
+using namespace std;
+
+/*
+Approach:
+1. The maximum absolute difference between any two elements in the array
+   is always (maximum element - minimum element).
+2. A pair (ai, aj) is interesting only if one of them is the minimum value
+   and the other is the maximum value.
+3. Count how many times the minimum value appears (cntMin)
+   and how many times the maximum value appears (cntMax).
+4. Since (ai, aj) and (aj, ai) are considered different pairs,
+   the total number of interesting pairs is:
+      2 * cntMin * cntMax
+5. Special case:
+   - If all elements are equal (min == max), then every ordered pair
+     (i, j) with i ≠ j is valid.
+   - The answer becomes n * (n - 1).
+*/
+
+int main(){
+
+    int t;
+    cin >> t;
+    while(t--){
+        int n;
+        cin >> n;
+
+        vector<long long> a(n);
+        for(int i = 0; i < n; i++) cin >> a[i];
+
+        long long mn = *min_element(a.begin(), a.end());
+        long long mx = *max_element(a.begin(), a.end());
+
+        if(mn == mx){
+            cout << 1LL * n * (n - 1) << "\n";
+            continue;
+        }
+
+        long long cntMin = 0, cntMax = 0;
+        for(long long x : a){
+            if(x == mn) cntMin++;
+            if(x == mx) cntMax++;
+        }
+
+        cout << 2LL * cntMin * cntMax << "\n";
+    }
+    return 0;
+}

Problems/Mathematics/Day-03/sol/Sujal_Kshatri/day03sol2.cpp

@@ -0,0 +1,35 @@
+// submission link - https://codeforces.com/contest/343/submission/355607613
+
+#include <bits/stdc++.h>
+using namespace std;
+
+/*
+1. We want to make fraction a/b using smallest number of 1 resistors.
+2. We can add resistor in line (series) or side (parallel).
+3. Instead of making from 1, we go backward from a/b.
+4. If a is bigger than b, then we minus b from a.
+5. If b is bigger than a, then we minus a from b.
+6. Every minus means we used one resistor.
+7. We count how many times we can minus using divide.
+8. This work same like Euclid algo (gcd thing).
+9. When one number become zero, we stop.
+10. Total count is our answer.
+*/
+
+int main()
+{
+    unsigned long long a, b;
+    cin >> a >> b;
+
+    long long totalResistors = 0;
+
+    while (a && b)
+    {
+        totalResistors += a / b;
+        a %= b;
+        swap(a, b);
+    }
+
+    cout << totalResistors;
+    return 0;
+}