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**HackerEarth The savior? [3] problem solution**Fatal Eagle is trying his best to save his city, but he's failing to do so. He knows he needs some external help; some guidance, something... something out of ordinary to defeat this powerful enemy called Mr. XYZ."Find out the number of pairs in the list you've made, who sum up to an even number!"

"Who are you? And what do you mean? Can you elaborate a bit?" Exclaims Fatal Eagle.

"I meant, what I said... take a look at the sample explanation of the problem I want you to solve, you'll understand perhaps. And as for who I am? I'm Arjit and like you, I just want to save this city, too."

Fatal Eagle is quick to respond to Arjit's task, and decides to solve it quickly. He takes out the list of integers which contains the powers of weird creatures, and starts finding out the number of pairs which sum up to an even number. Here's how he does it:

Let's say that he had a list of 4 numbers: [1, 2, 3, 4] His answer would be: 2. How? 1+3 = 4 and 2 + 4 - that is to say, a number will NOT be added to itself and 1+3 is same as 3+1.

## HackerEarth The savior? 3 problem solution.

`tc = int(raw_input())`

while tc>0:

tc = tc - 1

n = int(raw_input())

l = map(int, raw_input().split())

ans = 0

for i in xrange(n):

for j in xrange(n):

if l[i]==l[j]:

pass

else:

if (l[i]+l[j])%2==0:

ans = ans + 1

print ans/2

### Second solution

`#include <bits/stdc++.h>`

void assert_digit() {char c = getchar(); assert('0' <= c && c <= '9'); ungetc(c, stdin);}

using namespace std;

int N;

int main()

{

assert_digit();

int TEST;

scanf("%d", &TEST);

assert(getchar()=='\n');

assert(1<=TEST && TEST<=50);

while(TEST--)

{

assert_digit();

scanf("%d", &N);

assert(getchar()=='\n');

assert(1<=N && N<=1000);

int cnt[2]={0, 0};

unordered_map<int, int> cnt2;

for(int i=0; i<N; i++)

{

int a;

assert_digit();

scanf("%d", &a);

if(i!=N-1)

assert(getchar()==' ');

else if(TEST!=0)

assert(getchar()=='\n');

assert(1<=a && a<=100000);

cnt[a%2]++;

cnt2[a]++;

}

int ans=0;

for(int i=0; i<2; i++)

ans+=cnt[i]*(cnt[i]-1)/2;

for(auto& it: cnt2)

ans-=it.second*(it.second-1)/2;

printf("%d\n", ans);

}

assert(getchar()==EOF);

return 0;

}

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