lightoj1182

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 numberof one's is odd. Otherwise we say n has even parity. 21 = (10101)₂has odd parity since the number of one's is 3. 6 = (110)₂has even parity.

Now you are given n, we have to say whether nhas 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).

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
using namespace std;
int main(){
	int i,j,k,m,n,T,ans;
	scanf("%d",&T);
	for(i=1;i<=T;i++){
		scanf("%d",&n);
		ans=0;
		while(n!=0){
			if(n%2==1)ans++;
			n/=2;
		}
		if(ans%2==1)printf("Case %d: odd\n",i);
		else printf("Case %d: even\n",i);
	}
	return 0;罗旅洲
}

Output

For each case, print the case number and 'odd' if nhas odd parity, otherwise print 'even'.

二进制下1的个数。odd:奇数，even偶数