lightoj-1182

1182 - Parity
PDF (English) Statistics Forum
Time Limit: 0.5 second(s) Memory Limit: 32 MB
Given an integer n, first we represent it in binary. Then we count the number of ones. We say n has odd parity if the number of one's is odd. Otherwise we say n has even parity. 21 = (10101)2 has odd parity since the number of one's is 3. 6 = (110)2 has even parity.

Now you are given n, we have to say whether n has even or odd parity.

Input
Input starts with an integer T (≤ 1000), denoting the number of test cases.

Each case contains an integer n (1 ≤ n < 231).

Output
For each case, print the case number and 'odd' if n has odd parity, otherwise print 'even'.

Sample Input
Output for Sample Input
2
21
6
Case 1: odd
Case 2: even

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;

int counter(int num){
    
    int a = 0x55555555;
    int b = 0x33333333;
    int c = 0x0f0f0f0f;
    int d = 0x00ff00ff;
    int e = 0x0000ffff;
    num = (num&a)+((num>>1)&a);
    num = (num&b)+((num>>2)&b);
    num = (num&c)+((num>>4)&c);
    num = (num&d)+((num>>8)&d);
    num = (num&e)+((num>>16)&e);
    return num;
}

int main(){
    
    int T,n;
    scanf("%d",&T);
    for(int t=1;t<=T;t++){
        scanf("%d",&n);
        printf("Case %d: ",t);
        printf(counter(n)%2==0?"even
":"odd
");
    }
    
    return 0;
}