Description
The binomial coefficient C(n, k) has been extensively studied for its importance in combinatorics. Binomial coefficients can be recursively defined as follows:
C(n, 0) = C(n, n) = 1 for all n > 0;
C(n, k) = C(n − 1, k − 1) + C(n − 1, k) for all 0 < k < n.
Given n and k, you are to determine the parity of C(n, k).
Input
The input contains multiple test cases. Each test case consists of a pair of integers n and k (0 ≤ k ≤ n < 2^31, n > 0) on a separate line.
End of file (EOF) indicates the end of input.
Output
For each test case, output one line containing either a “0” or a “1”, which is the remainder of C(n, k) divided by two.
Sample Input
Raw
1 1
1 0
2 1
Sample Output
Raw
1
1
0
题意:给一个组合数C(n,k),判断是奇数还是偶数。
解题思路:C(n,k)(n>=k) N&k==k 为奇数,否则为偶数.
#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
long long n,k;//不用long long要mle
while(scanf("%lld%lld",&n,&k)!=EOF){
if((n&k)==k)
printf("1\n");
else
printf("0\n");
}
return 0;
}