POJ题目1430 Binary Stirling Numbers（数学，斯特林）

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本博客介绍如何计算斯特林数的第二类的奇偶性，通过给定的两个整数n和m，计算S(n,m)模2的结果。

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Binary Stirling Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 1828		Accepted: 706

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

Source

Central Europe 2001

判断斯特林数是否能整数2

ac代码

#include<cstdio>
#define lint __int64

lint getTwo ( lint x )
{
    lint cnt = 0, bit = 2;
    while ( x / bit )
    {
        cnt += x / bit;
        bit <<= 1;
    }
    return cnt;
}

int main()
{
    int d, n, k;
    scanf("%d",&d);
    while ( d-- )
    {
        scanf("%d%d",&n, &k);
        lint z = n - (k+2) / 2;
        lint w = (k-1) / 2;
        if ( getTwo(z) - getTwo(w) - getTwo(z-w) > 0 )
            printf("0\n");
        else printf("1\n");
    }
    return 0;

}