1430 Binary Stirling Numbers 判断第二类斯特林数的奇偶性 对于C(n,k),若n&k == k 则c(n,k)为奇数，否则为偶数。

Binary Stirling Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 1106		Accepted: 379

Description

The Stirling number of the second kind S(n, m) stands for the number of ways to partition a set of n things into m nonempty subsets. For example, there are seven ways to split a four-element set into two parts:

{1, 2, 3} U {4}, {1, 2, 4} U {3}, {1, 3, 4} U {2}, {2, 3, 4} U {1}

{1, 2} U {3, 4}, {1, 3} U {2, 4}, {1, 4} U {2, 3}.

There is a recurrence which allows to compute S(n, m) for all m and n.

S(0, 0) = 1; S(n, 0) = 0 for n > 0; S(0, m) = 0 for m > 0;

S(n, m) = m S(n - 1, m) + S(n - 1, m - 1), for n, m > 0.

Your task is much "easier". Given integers n and m satisfying 1 <= m <= n, compute the parity of S(n, m), i.e. S(n, m) mod 2.


Example

S(4, 2) mod 2 = 1.

Task

Write a program which for each data set:
reads two positive integers n and m,
computes S(n, m) mod 2,
writes the result.

Input

The first line of the input contains exactly one positive integer d equal to the number of data sets, 1 <= d <= 200. The data sets follow.

Line i + 1 contains the i-th data set - exactly two integers ni and mi separated by a single space, 1 <= mi <= ni <= 10^9.

Output

The output should consist of exactly d lines, one line for each data set. Line i, 1 <= i <= d, should contain 0 or 1, the value of S(ni, mi) mod 2.

Sample Input

1
4 2

Sample Output

1

#include<iostream>
#include<cstdio>
using namespace std;
int _count(int n,int k)
{
    int cnt=0;
    while(n) cnt+=n/k,n/=k;
    return cnt;
}
int main()
{
    int ci;scanf("%d",&ci);
    int n,k;
    while(ci--&&scanf("%d%d",&n,&k)==2)
    {
        int w = (k-1)/2;    //求的是floor((k-1)/2)的值，就是向下取整；
        int z = n - k + w; //求的是n - ceil((k+1)/2)的值，就是向上取整；
        int cnt=_count(z,2)-_count(w,2)-_count(z-w,2);
        printf("%d/n",cnt?0:1);
    }
    return 0;
}
/*
s（n，k）=c（z，w）（mod 2）
z=n-ceil（（k+1）/2）;   w=floor（（k-1）/2）;
*/

或者

#include<iostream>
#include<cstdio>
using namespace std;
int _count(int n,int k)
{
    int cnt=0;
    while(n) cnt+=n/k,n/=k;
    return cnt;
}
int main()
{
    int ci;scanf("%d",&ci);
    int n,k;
    while(ci--&&scanf("%d%d",&n,&k)==2)
    {
        int w = (k-1)/2;    //求的是floor((k-1)/2)的值，就是向下取整；
        int z = n - k + w; //求的是n - ceil((k+1)/2)的值，就是向上取整；
        int cnt=(z&w)==w;
        printf("%d/n",cnt?1:0);
    }
    return 0;
}
/*
s（n，k）=c（z，w）（mod 2）
z=n-ceil（（k+1）/2）;   w=floor（（k-1）/2）;
*/

对于C(n,k),若n&k == k 则c(n,k)为奇数，否则为偶数。