Merge pull request #1 from lightoj-dev/main

updating fork

1001/en.md

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+# LOJ 1001 - Opposite Task
+
+In this problem, you are given `T` test cases. Each test case contains an integer `n`, the value of which is `20` at max. Your task is to print out two non-negative integers `a` and `b`, such that `n = a + b`. Any values of `a` and `b` that satisfies the equation is accepted, however, neither of `a` nor `b` can be greater than `10`.
+
+It is one of the easiest problems on LightOJ volumes. However, some people might find it difficult to get an `Accepted` verdict. The reason could be one of the following:
+1. Printing a number greater than 10 
+2. Failing to provide proper output format
+3. Missing new line on each line
+
+If you are still stuck with this problem, check the codes below:
+
+### C
+-----
+```c
+#include <stdio.h>
+int main() {
+  int T, n;
+  scanf("%d", &T);
+  for (int i = 1; i <= T; i++) {
+    scanf("%d", &n);
+    if (n > 10) {
+      printf("10 %d\n", n - 10);
+    } else {
+      printf("0 %d\n", n);
+    }
+  }
+  return 0;
+}
+```
+
+### Java
+-----
+```java
+package com.loj.volume;
+
+import java.util.Scanner;
+
+class OppositeTask {
+  public static void main(String[] args) throws Exception {
+    Scanner scanner = new Scanner(System.in);
+    int T = scanner.nextInt(), n;
+    for (int i = 1; i <= T; i++) {
+      n = scanner.nextInt();
+      if (n > 10) {
+        System.out.printf("10 %d\n", n - 10);
+      } else {
+        System.out.printf("0 %d\n", n);
+      }
+    }
+  }
+}
+```

1007/en.md

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+# LOJ 1007 - Mathematically Hard
+
+_You have given `T` test cases and for each test case you have `a`, the lower bound, and `b`, the upper bound, find the summation of the
+scores of the numbers from a to b (inclusive), where <br>
+score (x) = n<sup>2</sup>,where n is the number of relatively prime numbers with x, which are smaller than x._
+
+---
+
+### Summary
+
+This is a number theory problem and for the `T` group of the input set, where each set consists of `a` and `b`, you have to find,square sum of all Euler function values from `a` to `b`.
+
+### Solution
+
+Before you get to the solution let's have some idea about `Euler's totient function` aka `phi`.<br>
+
+```
+Euler's totient function, also known as phi-function `ϕ(n)`, counts the number of integers between `1` and `n` inclusive, which are `coprime` to `n`. Two numbers are coprime if their greatest common divisor equals `1` (`1` is considered to be coprime to any number).
+```
+
+Have a look:
+
+- [wikipedia: Euler's totient function](https://en.wikipedia.org/wiki/Euler%27s_totient_function)
+- [cp-algorithms: Euler's totient function](https://cp-algorithms.com/algebra/phi-function.html)
+
+- [A quite useful cheat-sheet of CP: Phi Implementation](https://github.com/ar-pavel/CP-CheatSheet#631-euler-function)
+
+Using Euler's totient function, calculate number of coprime for all the numbers until max possible input. Now, let's assume you have `phi[]` array that contains all the `ϕ(n)` values. As you will need to print the sum of square values of the `ϕ(n)`, loop through the `phi[]` array and multiply them with own value (`phi[i] = phi[i]*phi[i]`). Now your `phi[]` array contains square values of each `ϕ(n)`. For each query, as you need to print `sum(phi[a] ... phi[b])`, you can optimize this process by generating a [cumulative sum array](https://www.tutorialspoint.com/cplusplus-program-for-range-sum-queries-without-updates) of of `phi[]`.
+
+The algorithm is:
+
+- find all Phi values from `1 - rangeMax`
+- calculate square value of each of them
+- generate cumulative sum array of squared phi values
+- for each query, print the `cumulative sum array[b]` - `cumulative sum array [a-1]`
+
+### C++
+-----
+```cpp
+#include <bits/stdc++.h>
+
+using namespace std;
+
+#define M 5000000
+
+int phi[M+2];
+unsigned long long phiSum[M+2];
+
+void calculatePhi(){
+    for(int i=2; i<=M; i++)
+        phi[i] = i;
+    for(int i =2; i<=M; i++)
+        if(phi[i]==i)
+            for(int j=i; j<=M; j+=i)
+                phi[j]-=phi[j]/i;
+}
+
+int main(){
+
+    calculatePhi();
+    phiSum[1] = 0;
+
+    for(int i=2; i<=M; i++)
+        phiSum[i]= ((unsigned long long)phi[i]* (unsigned long long)phi[i])+phiSum[i-1];
+
+    // for(int i=1; i<10; ++i)
+    //  cout<<phi[i]<<" "<<phiSum[i]<<endl;
+    // deb(phi[2]);
+    // deb(phiSum[20]);
+
+    int tc,t(1);
+    for(scanf("%d",&tc); tc; ++t,--tc){
+        int a,b;
+        scanf("%d%d",&a,&b);
+        unsigned long long x = phiSum[b]-phiSum[a-1];
+        printf("Case %d: %llu\n",t,x);
+    }
+    return 0;
+}
+```

1014/en.md

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+# 1014 - Ifter Party
+
+_You have given `T` test cases and for each test case you have `P`, number of piaju in the beginning, and `L` piaju left end of the day, find the value `Q`, the number of piaju's each of `C` contestant ate._
+
+---
+
+### Summary
+
+This is a number theory problem and for the `T` group of the input set, where each set consists of `P` and `L` number, you have to find all the divisors of piaju's eaten or `(P-L)`.
+
+### Solution
+
+Initially, you had `P` amount of piaju's, and end of the day it became `L`, so piaju's eaten by the contestants are `(P-L)`. As `C` contestants were invited and each of them ate `Q` piaju's each, `P - L = C * Q` stands true. So the result will be all possible divisors of `Q`, the number of piaju's each contestant ate.
+
+The algorithm is:
+
+- find the number of piaju's eaten
+- find all the unique divisors of the 'count of eaten piaju's'
+- sort and print all the divisors in ascending order that satisfy `(L<Q)` condition.
+- if no such solution exists print `impossible`.
+
+### Code
+
+#### C++
+
+```cpp
+    #include <bits/stdc++.h>
+
+    using namespace std;
+
+    int main(){
+        #ifndef ONLINE_JUDGE
+            freopen("in.txt","r",stdin);
+            freopen("out.txt","w",stdout);
+        #endif
+
+        int tc,t(1);
+        for(scanf("%d",&tc); tc; ++t,--tc){
+            long long piaju,left;
+
+            scanf("%lld%lld", &piaju,&left);
+
+            // exception
+            if(left*2>=piaju){
+                printf("Case %d: impossible\n",t);
+                continue;
+            }
+
+            printf("Case %d:",t);
+
+
+            long long piajuEaten = piaju-left;
+            // cout<<piajuEaten<<endl;
+
+            vector<long long> possibleValueOfQ;
+
+
+           // find all the divisors of piajuEaten
+            for(long long i=1; i*i<=piajuEaten; ++i)
+                if(piajuEaten%i==0){
+                    possibleValueOfQ.push_back(i);
+
+                if(piaju/i!=i)
+                    possibleValueOfQ.push_back(piajuEaten/i);
+                }
+
+            // sort the divisors to print the answer in ascending order
+            sort(possibleValueOfQ.begin(), possibleValueOfQ.end());
+
+            // print the divisors
+            for(auto x: possibleValueOfQ)
+                // if(x>left)
+                    printf(" %lld",x);
+
+
+            printf("\n");
+        }
+
+        return 0;
+    }
+
+```

1029/en.md

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+# LOJ 1029 - Civil and Evil Engineer
+
+### Problem Summary
+
+You have `n` houses `[1-n]` and a power station `0`. You are also given a set of wires, where every wire is in `u v w` format, meaning `connection from u to v costs w`. You have to use exactly `n` wires from the set to connect the houses with the power station. **The connections can be direct or indirect (through some houses) but there must be a path of wires from every house to the power station.**
+
+### Observations
+
+We can see the following properties in the problem structure.
+
+- **Graph**: We can consider the houses and the power station as `nodes` in a graph, and the wires as `edges`. Since the costs of connecting two nodes are given, and the connections have no direction, so this is a `Weighted Undirected Graph`.
+
+- **Tree**: In total, we have `n+1` nodes (`n` houses and `1` power station). We have to connect them using exactly `n` wires or `edges`. We know in a tree, for `N` nodes we have `N-1` edges. So the problem is asking us to convert the graph into a tree.
+
+- **Minimum Spanning Tree**: For the `best possible connection scheme`, We have to connect all houses with the power station using minimum cost. We have to pick `edges` in a way that satisfies the cost minimization. Therefore, we have to compute the `Minimum Spanning Tree` of the graph.
+
+  - `Minimum Spanning Tree` of a graph is a subtree of that graph that contains all the `nodes` and the `sum of edge weights` is minimum possible. You can use [Prim's algorithm](https://cp-algorithms.com/graph/mst_prim.html) or [Kruskal's Algorithm](https://cp-algorithms.com/graph/mst_kruskal.html) to implement the Minimum Spanning Tree.
+
+- **Maximum Spanning Tree**: We also have to calculate the maximum possible cost to connect all the `nodes`, since this is the `worst possible connection scheme`. For this, we have to pick `edges` in a way that maximizes the cost. So we also have to compute the `Maximum Spanning Tree` of the graph.
+  - `Maximum Spanning Tree` of a graph is a subtree of that graph that contains all the `nodes` and the `sum of edge weights` is maximum possible. We can tweak the Minimum Spanning Tree algorithm (sort the edges in descending order of the weights) to implement Maximum Spanning Tree.
+
+### Solution
+
+For every case, we have to calculate minimum spanning tree `cost1` and maximum spanning tree `cost2` and then average their costs, thus `(cost1 + cost2) / 2`. If `(cost1 + cost2)` is not divisible by `2` then we have to print in `p/q` format. Where `p` is `(cost1 + cost2)` and `q` is `2`. For example, `229/2`.
+
+### Simulation
+
+Simulation of **test case 2** is given below.
+
+![Frame 1](https://user-images.githubusercontent.com/14056189/99875765-3bdd5700-2c1c-11eb-970f-92222ebe7c22.png)
+
+![Frame 2](https://user-images.githubusercontent.com/14056189/99876684-8792ff00-2c22-11eb-8a0a-6fe3ee6bc73f.png)
+
+![Frame 3](https://user-images.githubusercontent.com/14056189/99875768-3f70de00-2c1c-11eb-80d2-38ab56236789.png)
+
+And the answer is `(70 + 159) / 2`. Since `229` is not divisible by `2`, we print `229/2`.
+
+### C++
+
+---
+
+```C++
+#include<bits/stdc++.h>
+using namespace std;
+
+struct edge{
+    int u, v, w;
+};
+
+bool cmp(edge A, edge B){
+    return A.w < B.w;
+}
+
+// edge list
+vector < edge > G;
+
+int n; // total nodes
+int parent[105];
+
+// function to clear graph
+void _init(){
+    G.clear();
+}
+
+// function to find parent of a disjoint set
+int Find(int u){
+    if (u==parent[u]) return u;
+    return parent[u] = Find(parent[u]);
+}
+
+// Kruskal's algorithm with Disjoint-Set Union method is used
+// minimum spanning tree
+long long MinST(){
+    // reset parent table [0-n]
+    for (int i = 0; i <= n; i++)
+        parent[i] = i;
+
+    long long cost = 0;
+    // forward iteration for minimum cost first
+    for (int i = 0; i < G.size(); i++){
+        int u = G[i].u;
+        int v = G[i].v;
+        int w = G[i].w;
+
+        int p = Find(u);
+        int q = Find(v);
+        if (p!=q){
+            cost += w;
+            parent[q] = p;
+        }
+    }
+    return cost;
+}
+
+// maximum spanning tree
+long long MaxST(){
+     // reset parent table [0-n]
+    for (int i = 0; i <= n; i++)
+        parent[i] = i;
+
+    long long cost = 0;
+    // backward iteration for maximum cost first
+    for (int i = G.size()-1; i >= 0; i--){
+        int u = G[i].u;
+        int v = G[i].v;
+        int w = G[i].w;
+
+        int p = Find(u);
+        int q = Find(v);
+        if (p!=q){
+            cost += w;
+            parent[q] = p;
+        }
+    }
+    return cost;
+}
+
+int main()
+{
+    int T; scanf("%d", &T);
+    for (int cs = 1; cs <= T; cs++){
+        _init(); // reset graph
+
+        scanf("%d", &n);
+        int u, v, w;
+
+        // take input until all u, v, w are zero (0)
+        while ( scanf("%d%d%d", &u, &v, &w) && (u!=0 || v!=0 || w!=0) ){
+            G.push_back({u, v, w});
+        }
+
+        // sorting is only done once
+        // forward and backward iterations will be done
+        // for ascending and descending order traversal
+        sort(G.begin(), G.end(), cmp);
+
+        long long cost1 = MinST();
+        long long cost2 = MaxST();
+
+        long long cost = cost1 + cost2;
+        if (cost%2 == 0) printf("Case %d: %lld\n", cs, cost/2);
+        else printf("Case %d: %lld/2\n", cs, cost); // cost/2 format
+    }
+    return 0;
+}
+```