【Ligth-oj】-1182 - Parity（水）

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 nhas 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

 

题意：就是看看一个数变成二进制有多少个1

#include<cstdio> #include<cstring>#include<algorithm>using namespace std;#define CLR(a,b) memset(a,b,sizeof(a))#define INF 0x3f3f3f3f#define LL long long#define PI acos(-1.0)int main(){int u,ca=1;scanf("%d",&u);while(u--){int n,num=0;scanf("%d",&n);while(n){if(n&1)num++;n>>=1;}if(num&1)printf("Case %d: odd\n",ca++);elseprintf("Case %d: even\n",ca++);}return 0;}