light-1182 - Parity【思维】

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1182 - Parity

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'.

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;
}