poj 1430 Binary Stirling Numbers 斯特灵数结论

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Binary Stirling Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 1914		Accepted: 756

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

Source

Central Europe 2001

题目大意：计算s{n,m}mod2的结果。

思路：可以利用结论（见下图）

ac代码：

#include<stdio.h>

int main()
{
    int t;
    int n,m;

    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        printf(((n-m)&((m-1)/2))==0?"1\n":"0\n");
    }
    return 0;
}

总结：网上还有别的思路可以看看。