## 100 - Office

Let's take into consideration a few cases:

- N % 4 == 0
- In this case, we can pair up complementary values (i and n-i+1) and give them the same sign. Because N is a multiple of 2
^{2}, the number of pairs is an even number (N/2 is a multiple of 2^{1}). We have reduced the problem to an even number of identical values, which can be given alternate signs, in order to add up to 0. In this case, the answer is 15.

- N % 4 == 2
- In this situation, we cannot apply the method described above, because we get one extra pair. The best we can do is assigning alternate signs from the very beginning (-1 +2 -3 +4 -5 +6, and so on). Each pair of two consecutive values gives either a +1 or a -1. Because there is an extra pair, we can choose to either have a final result of 14 (which is not valid), or 16 (which is the best we can do, considering that the sequence contains an odd quantity of odd numbers, thus it cannot evaluate to zero - an even number). In this case, the answer is 16.

- N % 4 == 3
- In this situation, we can group the first three elements in a fixed manner: +1 +2 -3 (adding up to zero). This reduces the problem to the last N-3 values. Considering that N-3 is a multiple of 4, we can apply the strategy described in the first case. The answer is 15.

- N % 4 == 1
- The same idea goes for this case: we leave the first element by itself (+1) and reduce the problem to N-1 values (which is also a multiple of four, so a zero sum can be obtained). Just like the second case, which had a lower bound of 1 for the number of extra flair pieces, the answer is 16.
# Python solution (Sergiu Puscas)

1 print 15 + [0, 1, 1, 0][input()%4]

## 250 - Coins

First of all you should observe that the problem specifies that you can beat the AI. This suggests that the player that begins the game can win.

It is possible there are more than one winning strategy, but the problem ask you to find only one. Now, if you dont't have any idea where to start the resource code might be a hint.

The winning strategy offered by us is the following:

- Put a coin in the center of the table (0,0)
- When the AI places a coin you place a coin symetricly to the center. (if he places a coin at (x,y), you will place it at (-x,-y)) Because all the AI moves are valid, the mirror moves will also be valid.

## 500 - Golf

Let `d[i]` be the number of ways we can split i*4 golfers into groups of 4 golfers each.

We have `d[1] = 1` and `d[i] = d[i - 1] * choose(i * 4, 4) / i`, where `choose(x, y)` is the binomial coefficient, calculated using the formula: `choose(x, y) = x! / y! / (x-y)!`.

When going from `d[i-1]` to `d[i]`, we can pick any 4 golfers to be in the newly formed group (`choose(i * 4, 4)`). However, we need to discount repetitions, thus need to divide by the number of groups i.

The final solution is d[N / 4]^{R}.

# Perl solution (Marius Gavrilescu)

In languages with builtin big numbers, we can compute d[N / 4] directly. Since d[N / 4]^{R}has more than 200 digits, we will print

`d[N / 4]^R`.

1 #!/usr/bin/perl 2 use v5.14; 3 use warnings; 4 use bigint; 5 6 my ($n, $r) = split ' ', <>; 7 my $ans = 1; 8 $ans *= (4 * $_)->bnok(4) / $_ for 2 .. $n / 4; 9 say "$ans^$r";

# C solution (Marius Gavrilescu)

We know that `choose(K,4)=K!/(K-4)!/4! = K*(K-1)*(K-2)*(K-3)/24`.

By expanding the definition of d we get `d[N/4] = N*(N-1)*(N-2)*(N-3) / 24 * (N-4)*(N-5)*(N-6)*(N-7) / 24 * ... * 4*3*2*1 / 24 / (N/4)!` which is equal to `N! / (N/4)! / 24^(N/4)`.

Thus, the answer is `(N*(N-1)*...*(N/4+1)/24^K)^R`.

1 #include <stdio.h> 2 3 int main(void) { 4 int n, k; 5 scanf("%d%d", &n, &k); 6 7 printf("(%d", n); 8 for(int i = n - 1 ; i > n / 4 ; i--) 9 printf("*%d", i); 10 printf("/24^%d)^%d\n", n / 4, k); 11 return 0; 12 }

# C++ solution (Gabriel Inelus)

We denote the number of ways to split N golfers into groups of 4 each when the order is important by the function`F(x) = x choose 4 * F(x-4)`. To avoid the repetitions we can divide the answer by

`(x/4)!`and then raise it to the power R. It's easy to observe that

`x choose 4 = x*(x-1)*(x-2)*(x-3)/24`, which is less than 2

^{31}, thus we can give a straight forward recursive implementation of F(N).

1 #include <bits/stdc++.h> 2 using namespace std; 3 4 void F(int k){ 5 cout << k*(k-1)*(k-2)*(k-3) / 24; 6 if(k == 4) return; 7 cout << "*"; 8 F(k - 4); 9 } 10 11 int main() 12 { 13 int N,R; 14 cin >> N >> R; 15 cout << "(("; F(N); cout << ")"; 16 for(int i = 2; i <= N/4; ++i) 17 cout << "/" << i; 18 cout << ")^" << R; 19 return 0; 20 }

## 1000 - Not another LCA problem

## Solution (Gabriel Robert Inelus)

First of all, we should start our solution by thinking about a way of counting the number of pairs ` {(v1,v2) | (v1,v2) != (v2,v1)}` where `LCA(v1,v2) = x ` and they have the property that `Value[v1] <= BigValue[x] and Value[v2] <= BigValue[x]`.

Considering the vertex x, the v1 and v2 vertices can be in two distinct subtrees denoted by the neighbours of vertex x. Let S be the total number of vertices from x's subtrees which have their Value less or equal to `BigValue[x]` excluding vertex x. The number of pairs `(v1,v2)` with `LCA(v1,v2) = x` and neither of v1 and v2 equal x is ` sum of S*(S-cnt(nb)) ` where we fix nb as every single neighbour of vertex x and `cnt(nb)` is the number of good vertices in nb's subtree including nb. Now, if `Value[x] <= BigValue[x]`, we have to consider `(v,x)` pairs where `LCA(v,x) = x` and symetrically, `(x,v)` pairs. Thus, in this case we have to add `2*S` pairs to the solution and also a unit which represents the pair `(x,x)`.

We can compute the solution for each vertex x in `O(logN)` time complexity if we consider that initially, neither of the vertices is good, and we consider each tuple of type `(Value[vertex], vertex, 1)` and `(BigValue[vertex],vertex,2)` as events. These events have to be incresingly sorted by the first element of each tuple and in case of equality, by the last element. Now, we can use a reduced imaginary eulerian representation of a rooted tree in this way: We build our imaginary eulerian path by taking into consideration a vertex when we enter the recursion only. Thus, we can use two vectors, `poz[x]` which denotes the position of the vertex x in the imaginary eulerian representation and `len[x]` the length of the continuous sequence of vertices which start on `poz[x]` and denotes the subtree rooted in x. We can easily observe that the interval `[poz[x]+1,poz[x]+len[x]-1]` represents the subtree of vertex x, excluding x. We can use a Fenwick Tree or a Segment Tree on the reduced eulerian path of the tree in order to have queries on this type of intervals to compute in `O(logN)` how many vertices of a subtree have Value less or equal to BigValue. We assume that initially, there is no good vertex. By iterating trough the events, if we find an event type 1, we have an update: at the position `poz[vertex]`, we should update from state 0 to state 1 and for type 2 events, we have to compute queries on the actual (updated) eulerian representation. Having the events sorted means that it is guaranteed that by making a query on a subtree, the sum on the interval which represents the specific subtree represents the the number that we are actually searching because all the good vertices in the whole three have already been marked with a 1.

Returning to our formula, when we reach a query event `(V,x,2)`, we compute S as a query on the interval `[poz[x]+1,poz[x]+len[x]-1]` and for each neighbour of x let nb denote them, we add to the current solution `S*(S-q)` where q is the result of a query on the interval `[poz[nb],poz[nb]+len[nb]+1]`. After this sum, if `Value[x] <= BigValue[x]`, we add to the current solution ` 2*S+1 ( or (S<<1)|1 )` which represent the rest of valid pairs.

The final complexity is `O(NlogN)` time and `O(N)` memory. Segment Trees compute the queries faster than Fenwick Trees.

## C++ solution (Cosmin Rusu)

1 #include <iostream> 2 #include <fstream> 3 #include <vector> 4 #include <algorithm> 5 6 using namespace std; 7 8 const int maxn = 100005; 9 10 int n, aib[maxn], in[maxn], out[maxn], k; 11 long long value[maxn], bigvalue[maxn], answer[maxn]; 12 vector <pair<long long, int> > events; 13 vector <int> g[maxn]; 14 15 inline void dfs(int node) { 16 in[node] = ++ k; 17 for(auto it : g[node]) 18 dfs(it); 19 out[node] = k; 20 } 21 22 inline int lsb(int x) { 23 return x & (-x); 24 } 25 26 inline void update(int pos, int value) { 27 for(int i = pos ; i <= n ; i += lsb(i)) 28 aib[i] += value; 29 } 30 31 inline int query(int pos) { 32 int sum = 0; 33 for(int i = pos ; i > 0 ; i -= lsb(i)) 34 sum += aib[i]; 35 return sum; 36 } 37 38 int main() { 39 cin >> n; 40 for(int i = 2 ; i <= n ; ++ i) { 41 int x; 42 cin >> x; 43 g[x].push_back(i); 44 } 45 dfs(1); 46 for(int i = 1 ; i <= n ; ++ i) { 47 cin >> value[i]; 48 events.push_back(make_pair(value[i], i)); 49 } 50 for(int i = 1 ; i <= n ; ++ i) { 51 cin >> bigvalue[i]; 52 events.push_back(make_pair(bigvalue[i], i + n)); 53 } 54 sort(events.begin(), events.end()); 55 for(auto event : events) { 56 int node = event.second; 57 if(node <= n) { // update 58 update(in[node], 1); 59 } 60 else { // query 61 node -= n; 62 long long ans = 0; 63 int sum = query(out[node]) - query(in[node]); 64 if(value[node] <= bigvalue[node]) { 65 ans += sum; // (node, sons) 66 ans += sum; // (sons, node) 67 ++ ans; // (node, node) 68 } 69 for(auto it : g[node]) { 70 int act = query(out[it]) - query(in[it] - 1); 71 ans += 1LL * act * (sum - act); 72 } 73 answer[node] = ans; 74 } 75 } 76 for(int i = 1 ; i <= n ; ++ i) 77 cout << answer[i] << '\n'; 78 }

## C++ solution (Gabriel Inelus)

1 #include <bits/stdc++.h> 2 #define Nmax 100005 3 4 using namespace std; 5 6 int N; 7 long long Value[Nmax],BigValue[Nmax]; 8 long long rasp[Nmax]; 9 int poz[Nmax]; 10 int len[Nmax]; 11 vector<int> G[Nmax]; 12 vector<pair<long long,int> > P; 13 14 void Read() 15 { 16 long long a; 17 scanf("%lld",&a); 18 N = a; 19 for(int i = 2; i <= N; ++i){ 20 scanf("%lld",&a); 21 G[a].push_back(i); 22 } 23 for(int i = 1; i <= N; ++i){ 24 scanf("%lld",&a); 25 Value[i] = a; 26 P.push_back(make_pair(a,i)); 27 } 28 for(int i = 1; i <= N; ++i){ 29 scanf("%lld",&a); 30 BigValue[i] = a; 31 P.push_back(make_pair(a,i+N)); 32 } 33 sort(P.begin(),P.end()); 34 } 35 int eup; 36 37 void DFS(int k){ 38 ++eup; 39 poz[k] = eup; 40 for(auto it : G[k]) 41 DFS(it); 42 len[k] = eup - poz[k] + 1; 43 } 44 45 int A,B,pos; 46 long long answer; 47 48 class SegmentTree{ 49 public: 50 vector<int> range; 51 void Resize(int k){ 52 range.resize(1 <<( (int)ceil(log2( (double) k)) + 1 ) ); 53 } 54 void Update(int li,int lf,int pz){ 55 if(li == lf){ 56 range[pz] = 1; 57 return; 58 } 59 int m = li + ((lf - li) >> 1); 60 if(pos <= m) Update(li,m,pz<<1); 61 else Update(m+1,lf,(pz<<1)|1); 62 range[pz] = range[pz<<1] + range[(pz<<1)|1]; 63 } 64 void Querry(int li,int lf,int pz){ 65 if(A <= li && lf <= B){ 66 answer += range[pz]; 67 return; 68 } 69 int m = li + ((lf - li) >> 1); 70 if(A <= m) Querry(li,m,pz<<1); 71 if(B > m) Querry(m+1,lf,(pz<<1)|1); 72 } 73 }; 74 SegmentTree Aint; 75 76 void Solve() 77 { 78 Aint.Resize(N); 79 int crt; 80 long long cst,S,vi; 81 for( auto it : P ) 82 { 83 crt = it.second; 84 cst = it.first; 85 if(crt <= N){ 86 pos = poz[crt]; 87 Aint.Update(1,N,1); 88 } 89 else 90 { 91 crt -= N; 92 answer = 0; 93 A = poz[crt] + 1; 94 B = poz[crt] + len[crt] - 1; 95 if(A <= B){ 96 Aint.Querry(1,N,1); 97 S = answer; 98 } 99 else 100 S = 0; 101 for(auto jt : G[crt]){ 102 answer = 0; 103 A = poz[jt]; 104 B = poz[jt] + len[jt] - 1; 105 Aint.Querry(1,N,1); 106 rasp[crt] += answer * (S - answer); 107 } 108 if(Value[crt] <= BigValue[crt]) 109 rasp[crt] += ((S<<1)|1); 110 } 111 } 112 for(int i = 1; i <= N; ++i) 113 printf("%lld\n",rasp[i]); 114 } 115 116 int main() 117 { 118 119 Read(); 120 DFS(1); 121 Solve(); 122 123 return 0; 124 }