Add Day 0 Question 1 solution

Author: priyanshigiri

Problems/Mathematics/Day-01/sol/Priyanshi_Giri/question1.cpp

@@ -0,0 +1,40 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+
+    int t;
+    cin >> t;
+    while (t--) {
+        int a, b, c;
+        cin >> a >> b >> c;
+        int n = 2*(abs(a-b));
+
+        if(a == b){
+            cout << -1 << endl;
+            continue;
+        }
+
+        if(a > n || b > n || c > n){
+            cout  << -1 << endl;
+            continue;
+        }
+
+        if(c + (n/2) <= n){
+            cout << (c+(n/2)) << endl;
+        }else{
+            cout << (c-(n/2)) << endl;
+        }
+
+    }
+    return 0;
+}
+
+
+
+
+/* Explaination : The opposite numbers must be having exactly n/2 numbers in between, hence the total number, ie. n = 2*abs(a-b).
+
+Time Complexity : O(t), t is the number of test cases. */