Merge pull request #116 from ayush-mg/main

Create Solution1.cpp

Author: ScorchedPearl

Problems/Mathematics/Day-01/sol/Ayush Mishra/Solution1.cpp

@@ -0,0 +1,43 @@
+// submission link
+// https://codeforces.com/contest/1560/submission/355180160
+
+// Space Complexity = O(1)
+// Time Complexity = O(1) for main solution code O(t) for overall code
+
+
+
+// The if condition checks if the circle is valid and if the inputs exist within that circle.
+// abs(b-a)<=1: Ensures a and b are not adjacent or the same.
+// 2*(abs(b-a)-1)+2 < c || 2*(abs(b-a)-1)+2 < a || 2*(abs(b-a)-1)+2 < b.  checks that the input numbers a, b, and c are not larger than the total size of the circle.
+// If any of these are true, it prints -1 .
+
+// The line int k = 2*(abs(b-a)-1) + 2 calculates the total number of items in the circle.
+
+// If the inputs are valid, it calculates the number opposite to c.
+// The halfway distance is k / 2.
+// If c is in the first half (c<= k/2), the opposite is c + k/2.
+// If c is in the second half (c > k/2), the opposite is c - k/2.
+
+
+
+
+#include<bits/stdc++.h>
+using namespace std;
+int main(){
+    int t;
+    cin>>t;
+    while(t--){
+        int a,b,c;
+        cin>>a>>b>>c;
+        if(abs(b-a)<=1 || 2*(abs(b-a)-1)+2 < c || 2*(abs(b-a)-1)+2 < a || 2*(abs(b-a)-1)+2 < b ){
+            cout<<-1<<"\n";
+        }
+        else {
+            int k =  2*(abs(b-a)-1) + 2;
+            if(c<=k/2) cout<<c+k/2 <<"\n";
+            else cout<<c-k/2<<"\n";
+        }
+
+    }
+    return 0;
+}