Merge branch 'main' of https://github.com/opencodeiiita/CP-Chronicles into day2-q001

Author: verifiedHuman18

.DS_Store

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+.vscode/ 

.gitignore

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+#include <bits/stdc++.h>
+using namespace std;
+/*
+Logic:
+The main key in solving the problem is that a,b will be enought to know that the size of the area.
+Another observation would be that 1 is always facing the element  more than the element which faces 
+at the last element or the size of the circle.
+Once, you get the size of the array, we just have to check if the any of the elements are larger than 
+the size of the array. 
+On deeper checks like- a=3, b=2,c=1;
+we can observe how this isn't true but how does the conditon fulfill the case?
+--> the max check covers all small checks, like the circle don being possible
+in cases like a=1,b=6, c=100
+--> the circle condition is also checked by the max case. 
+
+So, after the checking, we need to see the answer, that can be easily given as the
+difference is same for all pairs of elements  modular nature of the circle( cool words that make sense)
+ 
+Time complexity-->
+
+The time complexity is O(1)
+The space complexity is O(1)
+
+Submission link- https://codeforces.com/contest/1560/submission/355188467
+*/
+//Soln->
+int main() {
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int t;
+    cin >> t;
+
+    while (t--) {
+        int a,b,c;
+        cin>>a;
+        cin>>b;
+        cin>>c;
+        int x=abs(b-a);
+    if(max({a,b,c})>(2*x))
+    cout<<-1<<endl;
+    else
+    {
+        if(c>x)
+        cout<<c-x<<endl;
+        else
+        cout<<c+x<<endl;
+    }
+    
+    }
+    return 0;
+}
+    

Problems/.DS_Store

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+/* 1 to n people are standing in a circle. n is even.... a is looking at b. 
+If given c is looking at some d person for given a,b then find that d. If such a combination
+of circle does not exist then print -1*/
+
+// Time complexity: O(1)
+// Space complexity: O(1)
+//My submission: https://codeforces.com/contest/1560/submission/355204246
+
+#include <bits/stdc++.h>
+using namespace std;
+
+using ll = long long;
+#define all(x) (x).begin(), (x).end()
+
+const int MOD = 1e9 + 7; // 998244353
+const ll INF = 1e18;
+
+void solve() {
+    int a,b,c;
+    cin>>a>>b>>c;
+    int n=2*abs(a-b);
+    if(a>n || b>n ||c>n ) cout<<-1<<endl;
+    else{
+        int x=c-n/2,y=c+n/2;
+        if(x>=1 && x<=n) cout<<x<<endl;
+        else cout<<y<<endl;
+    }
+}
+
+int main() {
+    // Fast I/O
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int t;
+    cin >> t;
+    while (t--) {
+        solve();
+    }
+    
+    return 0;
+}

Problems/Mathematics/.DS_Store

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+//https://codeforces.com/contest/1560/problem/C
+//https://codeforces.com/contest/1560/submission/355210608
+// TC : O(1) SC: O(1)
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+	int t;
+	cin>>t;
+	while(t--){
+	    int a,b,c;
+	    cin>>a>>b>>c;
+	    int k=abs(a-b);
+	    if(a>2*k || b>2*k || c>2*k){
+	        cout<<-1<<endl;
+	    }
+	    else{
+	        if((c-k)>0){
+	            cout<<c-k<<endl;
+	        }
+	        else{
+	            cout<<c+k<<endl;
+	        }
+	    }
+	}
+
+}

Problems/Mathematics/Day-01/.DS_Store

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+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    int t;
+    cin >> t;
+
+    while (t--) {
+        int a, b, c;
+        cin >> a >> b >> c;
+
+        int half = abs(a - b);
+        int n = 2 * half;
+
+        if (c > n) {
+            cout << -1 << "\n";
+        } else {
+            int opposite = (c + half <= n) ? (c + half) : (c - half);
+            cout << opposite << "\n";
+        }
+    }
+    return 0;
+}