poj Binomial Coefficients （Lucas定理）

Binomial Coefficients

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 6819		Accepted: 2844

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 < 231,
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

1 1
1 0
2 1

Sample Output

1
1
0

Source
//题意：
给出C(n,k)的计算规则：
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.
要判断C（n，k）是否模2为1（即C（n，k）为奇数），由Lucas定理可知，C（n，k）mod 2=C（a[m],b[m]）×C（a[m-1],b[m-1）×...×C（a[0],b[0]）mod 2，其中，a[m]a[m-1]...a[0]，b[m]b[m-1]...b[0]分别为n，k的二进制表示。所以我们只需保证
n二进制位上为0的点 对应的 k二进制位上的该位置的值 不为1即可（C（0,1）=0，C（0,0）=C（1,0）=C（1,1）=1）。注意到0&1=0，0&0=0，1&0=0，1&1=1，即，只有存在C（0,1）时，n和k做位与运算后得到的值才和k不相等，那么如果不存在C（0,1），即C（n，k）为奇数时，n&k是等于k的。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<algorithm>
#include<iostream>
#define INF 0x3f3f3f3f
#define ull unsigned long long
#define ll long long
#define IN __int64
#define N 110
#define M 1000000007
using namespace std;
int main()
{
ll n,k;
while(scanf("%lld%lld",&n,&k)!=EOF)
{
if((n&k)==k)
printf("1\n");
else
printf("0\n");
}
return 0;
}