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[][];
 int cnm[][];
 void init()
 {
     int i,j;
     dp[][]=;
     for(i=;i<=;i++) dp[i][]=;
     for(i=;i<=;i++)
         for(j=;j<=i;j++)
             dp[i][j]=dp[i-][j-]+dp[i-][j]*j;
     for(i=;i<=;i++)
     {
         for(j=;j<=i;j++)
             printf("%d ",(dp[i][j]&));
         printf("\n");
     }

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

ac代码

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

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