LightOJ 1182 Parity

1182 - Parity

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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

 

把一个数改成2进制，求出里面1的个数，若是奇数输出odd

#include<cstdio>int main(){int t,k=0;scanf("%d",&t);while(t--){int n,num=0;scanf("%d",&n);while(n){if(n % 2 == 1)num++;n /= 2;}printf("Case %d: ",++k);if(num % 2 != 0)printf("odd\n");elseprintf("even\n");}}