POJ1430 Binary Stirling Numbers

@(POJ)[Stirling數, 排列組合, 數形結合]

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

Solution

題意:

求斯特林數$$ \left{ \begin{array}{} n \ k \end{array}{} \right} % 2$$$$n, m \in [1, 10^9]$$

這題直接求解肯定是會T的, 因此考慮優化.

轉載自sdchr博客

侵刪

代碼附上:

#include<cstdio>
#include<cctype>
using namespace std;

inline int read()
{
	int x = 0, flag = 1;
	char c;
	while(! isdigit(c = getchar()))
		if(c == '-')
			flag *= - 1;
	while(isdigit(c))
		x = x * 10 + c - '0', c = getchar();
	return x * flag;
}

void println(int x)
{
	if(x < 0)
		putchar('-'), x *= - 1;
	if(x == 0)
		putchar('0');
	int ans[1 << 5], top = 0;
	while(x)
		ans[top ++] = x % 10, x /= 10;
	for(; top; top --)
		putchar(ans[top - 1] + '0');
	putchar('\n');
}

long long getQuantity(int x)
{
	long long ret = 0;

	for(int i = 2; i <= x; i <<= 1)
		ret += x / i;

	return ret;
}

int calculate(int x, int y)
{
	return getQuantity(x) - getQuantity(y) - getQuantity(x - y) == 0;
}

int main()
{
	int T = read();

	while(T --)
	{
		int n = read(), m = read();
		int d = n - m, oddQua = (m + 1) / 2;
		println(calculate(d + oddQua - 1, oddQua - 1));
	}
}