Merge branch 'opencodeiiita:main' into main

Author: Nitinkr03

Problems/Mathematics/Day-01/sol/Abhay/Q1sol.cpp

@@ -0,0 +1,53 @@
+#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/Mathematics/Day-01/sol/BEESA_MANISH/Solution2.cpp

@@ -0,0 +1,76 @@
+/*
+n = number of elements (≤ 100,000)
+
+A = maximum value of ai (≤ 100,000)
+
+Space Complexity - 𝑂(𝐴+𝑛)
+
+Time Complexity - O(nlogA)​
+
+
+*/
+
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int n, k;
+    cin >> n >> k;
+
+    vector<int> a(n);
+    for (int i = 0; i < n; i++) cin >> a[i];
+
+    // Smallest Prime Factor sieve
+    const int MAXA = 100000;
+    vector<int> spf(MAXA + 1);
+    for (int i = 1; i <= MAXA; i++) spf[i] = i;
+    for (int i = 2; i * i <= MAXA; i++) {
+        if (spf[i] == i) {
+            for (int j = i * i; j <= MAXA; j += i)
+                if (spf[j] == j)
+                    spf[j] = i;
+        }
+    }
+
+    map<vector<pair<int,int>>, long long> freq;
+    long long ans = 0;
+
+    for (int x : a) {
+        map<int,int> cnt;
+
+        // factorization
+        while (x > 1) {
+            int p = spf[x];
+            int c = 0;
+            while (x % p == 0) {
+                x /= p;
+                c++;
+            }
+            cnt[p] += c;
+        }
+
+        vector<pair<int,int>> sig, need;
+
+        for (auto &it : cnt) {
+            int p = it.first;
+            int e = it.second % k;
+            if (e != 0) {
+                sig.push_back({p, e});
+                need.push_back({p, (k - e) % k});
+            }
+        }
+
+        sort(sig.begin(), sig.end());
+        sort(need.begin(), need.end());
+
+        ans += freq[need];
+        freq[sig]++;
+    }
+
+    cout << ans << "\n";
+    return 0;
+}

Problems/Mathematics/Day-01/sol/Coldesy/Coldesy.cpp

@@ -0,0 +1,38 @@
+//https://codeforces.com/contest/1560/submission/355205808
+//time complexity:O(t)
+//space complexity:O(1)
+#include <iostream>
+using namespace std;
+
+int main()
+{
+    int t;
+    cin>>t;
+    while(t--){
+        int a,b,c;
+        cin>>a>>b>>c;
+        int len;
+        if(a>b){
+            len=a-b;
+            len=2*len;
+        }               // staring at opposite end means both of them divide the circle 
+                        // in exact half so we substract them and multiply by 2 to get exact length
+        else {
+           
+            len=b-a;
+            len=2*len;
+        
+        }
+        if(a<=len&&b<=len&&c<=len){//a,b,c ofcourse cannot be greater than length
+            if(c+(len/2)>len){
+                cout<<c-(len/2)<<endl;//c will either stare at +n/2 or -n/2
+            }
+            else{
+                cout<<c+(len/2)<<endl;
+            }
+        }
+        else{cout<<-1<<endl;}
+    }
+
+    return 0;
+}

Problems/Mathematics/Day-01/sol/Harshit Sethi/Solution1.cpp

@@ -0,0 +1,22 @@
+// https://codeforces.com/problemset/submission/1560/355220642
+
+#include <iostream>
+#include <cmath>
+using namespace std;
+int main()
+{
+    int t;
+    cin >> t;
+    while (t--)
+    {
+        long long a, b, c;
+        cin >> a >> b >> c;
+        long long order = fabs(a - b);
+        if ((2 * order < a) || (2 * order < b) || (2 * order < c))
+            cout << -1 << endl;
+        else if (order < c)
+            cout << c - order << endl;
+        else
+            cout << c + order << endl;
+    }
+}

Problems/Mathematics/Day-01/sol/Himansh_Arora/Solution1.cpp

@@ -0,0 +1,25 @@
+//https://codeforces.com/contest/1560/submission/355194892
+
+#include <bits/stdc++.h>
+using namespace std;
+
+void solve(){
+    int a,b,c;
+    cin>>a>>b>>c;
+    int n=2*abs(a-b);
+
+    int opp=c+n/2;
+    if(c>n||b>n||a>n){
+        cout<<-1<<endl;
+        return;
+    }
+    if(opp>n)opp=opp-n;
+    
+    cout<<opp<<endl;
+}
+
+int main(){
+    int t;
+    cin>>t;
+    while(t--)solve();
+}