Doubloon Game HDU-4203 Game Theory, NP Relational Reasoning

• First enumerate $k$ , first enumerate when$When\; k$ is an odd number, when$k=At\; 1$ , obviously$When\; n$ is an odd number, the first mover will win. At this time, you need to take 1 step to get to$P$ point.
• When $k=At\; 3$ :

Subscript	0	1	2	3	4	5	6	7	8	9
status	$P$	$N$	$P$	$N$	$P$	$N$	$P$	$N$	$P$	$N$

• Similarly when $k=At\; 5$ , the status table is the same as the above figure; it can be found that when$When\; k$ is odd,nn isalways satisfied$When\; n$ is odd, the first mover will win, and one step can lead to$P$ point; vice versa$When\; n$ is an even number, the first move is defeated.
• When $When\; k$ is even, draw$k=$The figure of$2$ is as follows:

Subscript	0	1	2	3	4	5	6	7	8	9
status	$P$	$N$	$N$	$P$	$N$	$N$	$P$	$N$	$N$	$P$

• Similarly draw when $k=$The state diagram at$4$ is as follows:

Subscript	0	1	2	3	4	5	6	7	8	9	10	11
status	$P$	$N$	$P$	$N$	$N$	$P$	$N$	$P$	$N$	$N$	$P$	$N$

• It can be found that when $When\; k$ is an even number, start from 0,$k+1$ length is a cycle, where$n\%(k+1)=When\; k$ is$N$ points, the remaining$0to(k-1)$ In the case of odd numbers.
• So when $n\%(k+1)=When\; k$ , you need to walk$k$ steps to the next$P$ point, otherwise as$n\%(k+1)\; When\; it$ is an odd number, one step can lead to the next$P$ point.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uii;
typedef pair<int, ll> pii;
template<typename T>
inline void rd(T& x)
{
    
    
	int tmp = 1; char c = getchar(); x = 0;
	while (c > '9' || c < '0') {
    
     if (c == '-')tmp = -1; c = getchar(); }
	while (c >= '0' && c <= '9') {
    
     x = x * 10 + c - '0'; c = getchar(); }
	x *= tmp;
}
const int N = 2e5 + 10;
const int M = 1e7 + 10;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
int n, m, k;
int main() {
    
    
#ifdef _DEBUG
	FILE* _INPUT = freopen("input.txt", "r", stdin);
	//	FILE* _OUTPUT = freopen("output.txt", "w", stdout);
#endif // !_DEBUG
	int cas = 0, T = 0;
	rd(T);
	while (T--) {
    
    
		//	while (~scanf("%d %d", &n,&k)) {
    
    
		rd(n), rd(k);
		if (k & 1) {
    
    
			if (n & 1) puts("1");
			else puts("0");
		}
		else {
    
    
			int tmp = n % (k + 1);
			if (tmp == k) printf("%d\n", k);
			else if (tmp & 1) puts("1");
			else puts("0");
		}
	}
	return 0;
}