# B. Button Pairs:

Tester: Taranpreet Singh (taran_1407)

SIMPLE

Casework, Math

# PROBLEM:

Given an array A of length N, find the number of pairs of numbers with an odd sum.

Subtask 1 [15 points]: N \leq 2
Subtask 2 [45 points]: N \leq 1000
Subtask 3 [40 points]: N \leq 100000

# QUICK EXPLANATION:

Each pair must have an odd number and an even number, so output the product of the number of odd numbers and the number of even numbers.

# EXPLANATION:

For this subtask, we note that N \leq 2. This means that the array has size 1 or 2. Letâ€™s consider the two cases:

If the array has size 1, then there are no pairs we can select. As a result, the answer must be 0.

Otherwise, if the array has size 2, then there is only one possible pair. We check if the sum of the two numbers in the array is odd. If it is, we output 1. If it is not, we output 0.

For this subtask, we note that N \leq 1000. This means that the array has size at most 1000, and a solution running in O(N^2) should pass. To solve this subtask, we use a naive solution. We maintain a counter of the number of pairs with an odd sum, initialised at 0.

We loop through every possible pair in the input, using two nested for loops. If the pair has an odd sum, we add one to our counter. At the end, we output the value of our counter.

In the final subtask, we note that N \leq 100000. Our O(N^2) solution will probably not pass for the subtask. For this, we attempt to do a little bit of casework.

We consider the different possibilities for the pair of numbers. Specifically, we note the parities of every combination:

• If we add an even number to an even number, we get an even number.
• If we add an odd number to an odd number, we get an even number.
• If we add an odd number to an even number, we get an odd number.

In particular, we notice that only the third case is important, when we add one odd number and one even number. We must find the number of occurrences of the third case.

We see that the pairs must contain one even and one odd number. In particular, if there are E even numbers in the array, and O odd numbers in the array, then the number of valid pairs will be E * O (there are E possible values of the even number in the pair and O possible values of the odd number). This solution obtains a perfect score for this task.

# SOLUTIONS:

Setter's Solution
#include "bits/stdc++.h"
using namespace std;

void solve() {

long long n;
cin >> n;
long long odd = 0;
long long even = 0;
for (long long i=0; i<n; i++) {
long long x;
cin >> x;
if (x % 2 == 0) even++;
else odd++;
}

cout << odd * even << endl;

}

int main() {

int t;
cin >> t;
while (t--) solve();

}

Tester's Solution
    import java.math.BigInteger;
import java.util.*;
import java.io.*;
import java.text.*;
public class Main{
//SOLUTION BEGIN
//Into the Hardware Mode
void pre() throws Exception{}
void solve(int TC)throws Exception {
int n = ni();
int[] c = new int[2];
for(int i = 0; i< n; i++)c[ni()%2]++;
pn(c[0]*(long)c[1]);
}
//SOLUTION END
void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
void exit(boolean b){if(!b)System.exit(0);}
long IINF = (long)1e15;
final int INF = (int)1e9+2, MX = (int)2e6+5;
DecimalFormat df = new DecimalFormat("0.00000000000");
double PI = 3.141592653589793238462643383279502884197169399, eps = 1e-7;
static boolean multipleTC = true, memory = true, fileIO = false;
void run() throws Exception{
if(fileIO){
out = new PrintWriter("");
}else {
out = new PrintWriter(System.out);
}
//Solution Credits: Taranpreet Singh
int T = (multipleTC)?ni():1;
pre();
for(int t = 1; t<= T; t++)solve(t);
out.flush();
out.close();
}
public static void main(String[] args) throws Exception{
if(memory)new Thread(null, new Runnable() {public void run(){try{new Main().run();}catch(Exception e){e.printStackTrace();}}}, "1", 1 << 28).start();
else new Main().run();
}
int find(int[] set, int u){return set[u] = (set[u] == u?u:find(set, set[u]));}
int digit(long s){int ans = 0;while(s>0){s/=10;ans++;}return ans;}
long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
void p(Object o){out.print(o);}
void pn(Object o){out.println(o);}
void pni(Object o){out.println(o);out.flush();}
String n()throws Exception{return in.next();}
String nln()throws Exception{return in.nextLine();}
int ni()throws Exception{return Integer.parseInt(in.next());}
long nl()throws Exception{return Long.parseLong(in.next());}
double nd()throws Exception{return Double.parseDouble(in.next());}

StringTokenizer st;
}

}

String next() throws Exception{
while (st == null || !st.hasMoreElements()){
try{
}catch (IOException  e){
throw new Exception(e.toString());
}
}
return st.nextToken();
}

String nextLine() throws Exception{
String str = "";
try{