Problem 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. 

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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

题意：t 组数据，每组给出第二类斯特林数的 S(n,m) 的 n、m，求其奇偶性

思路：

首先求 S(n,m) 的奇偶性实质就是求 S(n,m)%2

对于第二类斯特林数，首先有：S[i][j]=S[i−1][j−1]+j∗S[i−1][j]

那么在模 2 的情况下 ，有：

• 当 j 为偶数：S[i][j] ≡（S[i−1][j−1]%2）
• 当 j 为奇数：S[i][j] ≡（(S[i−1][j−1]+S[i−1][j])%2）

于是，可以倒过来，即有：

• 当 j 为偶数时：S[i][j] 会被加到 S[i+1][j+1] 和 S[i+1][j] 
• 当 j 为奇数时：S[i][j] 会被加到 S[i+1][j+1]

于是，令 a=n−m，b=(m+1)/2，则答案就相当于把 a 个相同的球分成 b 组，每组个数可以为 0 的方案总数，即：C(b−1，a+b−1)

然后利用 C(n，m) 为奇数时有 n&m=n 的结论，进行判断即可

最后，注意特判

Source Program

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define EPS 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LL long long
const int MOD = 1E9+7;
const int N = 10000+5;
const int dx[] = {0,0,-1,1,-1,-1,1,1};
const int dy[] = {-1,1,0,0,-1,1,-1,1};
using namespace std;

int calc(int n,int m){
    return (m&n)==n;
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        int n,m;
        scanf("%d%d",&n,&m);
        if(n==0&&m==0)//特判
            printf("1\n");
        else if(n==0||m==0||n<m)//特判
            printf("0\n");
        else{
            int a=n-m;
            int b=(m+1)/2;
            int res=calc(b-1,a+b-1);
            printf("%d\n",res);
        }
    }
    return 0;
}