POJ1430 ZOJ1385 UVA1118 UVALive2431 Binary Stirling Numbers【斯特林数】

Binary Stirling Numbers
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 1971 Accepted: 794

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

Regionals 2001 >> Europe - Central
Regionals 2001 >> Europe - Southwestern

问题链接：POJ1430 ZOJ1385 UVA1118 UVALive2431 Binary Stirling Numbers
问题简述：（略）
问题分析：
    这是一个Stirling Numbers（斯特林数）数问题。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* POJ1430 ZOJ1385 UVA1118 UVALive2431 Binary Stirling Numbers */

#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
    int d, n, m;
    scanf("%d", &d);
    for(int i = 0; i < d; i++) {
        scanf("%d %d", &n, &m);
        if((n - m) & (m - 1) / 2)
            printf("0\n");
        else
            printf("1\n");
    }

    return 0;
}