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
1 #include<iostream>
2 #include<stdio.h>
3 #include<cstring>
4 #include<cstdlib>
5 using namespace std;
6
7 int dp[21][21];
8 int cnm[21][21];
9 void init()
10 {
11 int i,j;
12 dp[0][0]=1;
13 for(i=1;i<=20;i++) dp[i][0]=0;
14 for(i=1;i<=20;i++)
15 for(j=1;j<=i;j++)
16 dp[i][j]=dp[i-1][j-1]+dp[i-1][j]*j;
17 for(i=0;i<=15;i++)
18 {
19 for(j=0;j<=i;j++)
20 printf("%d ",(dp[i][j]&1));
21 printf("\n");
22 }
23
24 cnm[0][0]=1;
25 for(i=1;i<=20;i++)
26 {
27 cnm[i][0]=1;
28 cnm[0][i]=1;
29 }
30 for(i=1;i<=20;i++)
31 {
32 for(j=1;j<=i;j++)
33 {
34 if(j==1) cnm[i][j]=i;
35 else if(i==j) cnm[i][j]=1;
36 else cnm[i][j]=cnm[i-1][j]+cnm[i-1][j-1];
37 }
38 }
39 for(i=0;i<=15;i++)
40 {
41 for(j=0;j<=i;j++)
42 printf("%d ",cnm[i][j]&1);
43 printf("\n");
44 }
45 }
46 int main()
47 {
48 init();
49 int n,m;
50 while(scanf("%d%d",&n,&m)>0)
51 {
52 printf("%d\n",dp[n][m]);
53 }
54 return 0;
55 }
ac代码
1 #include<iostream>
2 #include<stdio.h>
3 #include<cstring>
4 #include<cstdlib>
5 using namespace std;
6
7 int a[64],alen;
8 int b[64],blen;
9 void solve(int n,int m)
10 {
11 int i;
12 bool flag=false;
13 alen=0;
14 blen=0;
15 memset(a,0,sizeof(a));
16 memset(b,0,sizeof(b));
17 while(n)
18 {
19 a[++alen]=(n&1);
20 n=n>>1;
21 }
22 while(m)
23 {
24 b[++blen]=(m&1);
25 m=m>>1;
26 }
27 for(i=1; i<=alen; i++)
28 {
29 if(a[i]==0 && b[i]==1) flag=true;
30 if(flag==true) break;
31 }
32 if(flag==true)printf("0\n");
33 else printf("1\n");
34 }
35 int main()
36 {
37 int T;
38 int n,m,k;
39 scanf("%d",&T);
40 while(T--)
41 {
42 scanf("%d%d",&n,&m);
43 if(m%2==0)
44 {
45 n=n-1;
46 m=m-1;
47 }
48 k=(m+1)/2;
49 solve(n-k,m-k);
50 }
51 return 0;
52 }
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