light-1182 - Parity【思维】

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Somethingwll

最新推荐文章于 2025-08-08 20:13:30 发布

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本文介绍了一个简单的算法问题，即判断一个整数在二进制表示下的一的个数是奇数还是偶数。通过不断除以2并检查余数的方法来统计一的个数，最终确定其奇偶性。

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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

Case 1: odd

Case 2: even

题解：就是找二进制数有几个 1

#include<cstdio>
#include<algorithm>
#include<cstring>
#define LL long long
using namespace std;
LL n;
int main()
{
	int t,text=0;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lld",&n);
		LL sum=0,cnt=0; 
		while(n)
		{
			if(n%2==1)	cnt++;
			n/=2;
		}
		printf("Case %d: ",++text);
		if(cnt%2)	puts("odd");
		else	puts("even");
	}
	return 0;
}