1030: 牛人题目(位运算二)——Power OJ

最新推荐文章于 2022-01-13 17:13:44 发布

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最新推荐文章于 2022-01-13 17:13:44 发布

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本文介绍了一种快速判断组合数C(n,k)奇偶性的算法，通过位运算实现，避免了大数运算的复杂性。适用于编程竞赛和数学问题求解。

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;
}