poj 1430 Binary Stirling Numbers

Binary Stirling Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 1761		Accepted: 671

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

 

可以转化成求C（N,M)来做。当然不是直接转化。

打出表看一下，发现是有规律的。

每一列都会重复一次。打表看一下吧。

思路：  

　　　  s(n,m)   如果m是偶数  n=n-1; m=m-1;==>转化到它的上一个s(n-1,m-1);

          k=(m+1)/2;  n=n-k; m=m-k;求C（n,m)的奇偶性就可以了。（当然有很多书写方式，不一定要这样做。）

测试用的

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
using namespace std;

int dp[21][21];
int cnm[21][21];
void init()
{
    int i,j;
    dp[0][0]=1;
    for(i=1;i<=20;i++) dp[i][0]=0;
    for(i=1;i<=20;i++)
        for(j=1;j<=i;j++)
            dp[i][j]=dp[i-1][j-1]+dp[i-1][j]*j;
    for(i=0;i<=15;i++)
    {
        for(j=0;j<=i;j++)
            printf("%d ",(dp[i][j]&1));
        printf("
");
    }

    cnm[0][0]=1;
    for(i=1;i<=20;i++)
    {
        cnm[i][0]=1;
        cnm[0][i]=1;
    }
    for(i=1;i<=20;i++)
    {
        for(j=1;j<=i;j++)
        {
            if(j==1) cnm[i][j]=i;
            else if(i==j) cnm[i][j]=1;
            else cnm[i][j]=cnm[i-1][j]+cnm[i-1][j-1];
        }
    }
    for(i=0;i<=15;i++)
    {
        for(j=0;j<=i;j++)
            printf("%d ",cnm[i][j]&1);
        printf("
");
    }
}
int main()
{
    init();
    int n,m;
    while(scanf("%d%d",&n,&m)>0)
    {
        printf("%d
",dp[n][m]);
    }
    return 0;
}

ac代码

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
using namespace std;

int a[64],alen;
int b[64],blen;
void solve(int n,int m)
{
    int i;
    bool flag=false;
    alen=0;
    blen=0;
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    while(n)
    {
        a[++alen]=(n&1);
        n=n>>1;
    }
    while(m)
    {
        b[++blen]=(m&1);
        m=m>>1;
    }
    for(i=1; i<=alen; i++)
    {
        if(a[i]==0 && b[i]==1) flag=true;
        if(flag==true) break;
    }
    if(flag==true)printf("0
");
    else printf("1
");
}
int main()
{
    int T;
    int n,m,k;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        if(m%2==0)
        {
            n=n-1;
            m=m-1;
        }
        k=(m+1)/2;
        solve(n-k,m-k);
    }
    return 0;
}