codeforces A. Parity

A. Paritytime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given an integer nn

(n≥0n≥0

) represented with kk

digits in base (radix) bb

. So,n=a1⋅bk−1+a2⋅bk−2+…ak−1⋅b+ak.n=a1⋅bk−1+a2⋅bk−2+…ak−1⋅b+ak.

For example, if b=17,k=3b=17,k=3

and a=[11,15,7]a=[11,15,7]

then n=11⋅172+15⋅17+7=3179+255+7=3441n=11⋅172+15⋅17+7=3179+255+7=3441

.Determine whether nn

is even or odd.InputThe first line contains two integers bb

and kk

(2≤b≤1002≤b≤100

, 1≤k≤1051≤k≤105

) — the base of the number and the number of digits.The second line contains kk

integers a1,a2,…,aka1,a2,…,ak

(0≤ai<b0≤ai<b

) — the digits of nn

.The representation of nn

contains no unnecessary leading zero. That is, a1a1

can be equal to 00

only if k=1k=1

.OutputPrint “even” if nn

is even, otherwise print “odd”.You can print each letter in any case (upper or lower).ExamplesInputCopy13 3
3 2 7
OutputCopyeven
InputCopy10 9
1 2 3 4 5 6 7 8 9
OutputCopyodd
InputCopy99 5
32 92 85 74 4
OutputCopyodd
InputCopy2 2
1 0
OutputCopyeven
NoteIn the first example, n=3⋅132+2⋅13+7=540n=3⋅132+2⋅13+7=540

, which is even.In the second example, n=123456789n=123456789

is odd.In the third example, n=32⋅994+92⋅993+85⋅992+74⋅99+4=3164015155n=32⋅994+92⋅993+85⋅992+74⋅99+4=3164015155

is odd.In the fourth example n=2n=2

#include <iostream>
#include <bits/stdc++.h>
#include <string.h>
#include <math.h>
using namespace std;
long long int n[100010];
int main()
{long long int a,b,c,i,sum=0;cin >> a >> b;for(i=b; i>0; i--){cin >> c;if(a%2!=0){if(i==1){if(c%2!=0){n[i]=1;}else{n[i]=2;}}else{if(c%2!=0){n[i]=1;}else{n[i]=2;}}}else{if(i==1){if(c%2!=0){n[i]=1;}else{n[i]=2;}}else{n[i]=2;}}}for(i=b; i>0; i--){sum=sum+n[i];}if(sum%2==0){cout << "even";}else{cout << "odd";}return 0;
}

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