POJ 1385-Binary Stirling Numbers(判断第二类斯特林数的奇偶性)

来源：互联网发布：java 对象初始化编辑：程序博客网时间：2024/05/14 07:35

Binary Stirling Numbers

Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1430

Appoint description:

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

14 2

题意：
将n个连续的数分成m个非空的部分的分法：满足公式S(0,0)=1, S(n,0)=0, S(0,m)=0,S(n,m) = mS(n-1,m) + S(n-1, m-1), 求S(n,m)%2的值
分析：设f(n, m)=S(n,m)%2, 则当m为偶数时，f(n,m)=f(n-1,m-1), 当m为奇数时，f(n,m)=f(n-1,m)+f(n-1,m-1)。建立以n为x轴，以m为y轴，根据以上的公式算出从(0,0)到(n,m)的路径，得出f(n,m)=C((n-m), (m-1)/2),要求f(n,m)%2,根据C(n,m)=n!/(m!(n-m)!),若C(n,m)的分母中2因子的个数等于分子中2因子的个数，则有C(n,m)为奇数，否则C(n,m)为偶数

详细解析：点击打开链接

#include <stdio.h>#include <math.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <sstream>#include <algorithm>#include <set>#include <queue>#include <stack>#include <map>using namespace std;typedef long long LL;const int inf=0x3f3f3f3f;const double pi= acos(-1.0);#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1int cnt_2(int x)//x!中2因子的个数{    int cnt=0;    while(x){        x/=2;        cnt+=x;    }    return cnt;}int main(){    int T,n,m;    scanf("%d",&T);    while(T--){        scanf("%d %d",&n,&m);        n=n-m;        m=(m-1)/2;        if(cnt_2(n+m)==cnt_2(n)+cnt_2(m))//如果分子分母2的个数相等则为奇数            puts("1");        else//否则为偶数            puts("0");    }    return 0;}

1 0