题目链接:http://poj.org/problem?id=3219
Time Limit: 1000MS | Memory Limit: 131072K | |
Total Submissions: 6875 | Accepted: 2872 |
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
PKU Local 2007 (POJ Monthly--2007.04.28), Ikki
本来想通过record数组记录将各个n和k的可能组合数先生成出来,结果发现就算只用bool去存内存也会爆掉,后来查到这是一个非常经典的数学问题,意识到自己的数学是多么的渣啊。。。
根据杨辉三角的性质,组合数C(n, k)的奇偶性取决于k和n-k的二进制编码各位相与的结果是否为0,如果结果为0则组合数为奇数,如果非0则组合数为偶数。
#include <cstdio>
#include <cstdlib>
#include <iostream>
#define MAXN 2147483647
using namespace std;
int main() {
int n, k;
while (scanf("%d %d", &n, &k) != EOF) {
printf("%d\n", k&(n-k)?0:1);
}
return 0;
}