$$$a \oplus 1$$$ is equivalent to $$$a - 1$$$ when $$$a$$$ is an odd number, otherwise it is equivalent to $$$a + 1$$$.

If you apply $$$a \gets a \oplus 1$$$ when $$$a$$$ is odd, then that operation should make $$$a = b$$$.

#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
using namespace std;
int t, a, b, x, y;
int main(){scanf("%d", &t);
	while (t --){scanf("%d%d%d%d", &a, &b, &x, &y);
		if (a > b) printf("%d\n", (a ^ 1) == b ? y : -1);
		else {int c0 = b - a, c1 = (b + 1 >> 1) - (a + 1 >> 1);
		printf("%lld\n", y > x ? 1ll * c0 * x : 1ll * (c0 - c1) * x + 1ll * c1 * y);}
	}
}

Transform $$$(p_x,p_y)$$$ to $$$(q_x,q_y)$$$ into an operation where $$$a_i$$$ is the distance between the two points.

When $$$n+1=3$$$, there is a solution if and only if the largest $$$a_i$$$ is less than or equal to the sum of the remaining $$$a_i$$$.

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
#define N 100010
#define int long long
int t,n,sx,sy,tx,ty;
double a[N];
signed main()
{
	cin>>t;
	while(t--)
	{
		cin>>n;
		cin>>sx>>sy>>tx>>ty;
		for(int i=1;i<=n;i++)
		 cin>>a[i];
		a[++n]=sqrt((sx-tx)*(sx-tx)+(sy-ty)*(sy-ty));
		sort(a+1,a+n+1);
		double sum=a[n];
		for(int i=1;i<n;i++)
		 sum-=a[i];
		if(sum<=0)
		 puts("Yes");
		else
		 puts("No");
	}
}

#include<iostream>
#include<algorithm>
using namespace std;
#define N 100010
#define int long long
int t,n,sx,sy,tx,ty,a[N];
signed main()
{
	cin>>t;
	while(t--)
	{
		cin>>n;
		cin>>sx>>sy>>tx>>ty;
		for(int i=1;i<=n;i++)
		 cin>>a[i];
		sort(a+1,a+n+1);
		int p=(sx-tx)*(sx-tx)+(sy-ty)*(sy-ty);
		if(p>a[n]*a[n])
		{
			int sum=0;
			for(int i=1;i<=n;i++)
			{
				sum+=a[i];
				if(sum*sum>=p)
				 break;
			}
			if(sum*sum>=p)
			 puts("Yes");
			else
			 puts("No");
		}
		else
		{
			int sum=a[n];
			for(int i=1;i<=n-1;i++)
			 sum-=a[i];
			if(sum<=0||sum*sum<=p)
			 puts("Yes");
			else
			 puts("No");
		}
	}
}

For odd $$$n$$$, setting $$$a_i = l$$$ for all $$$i$$$ suffices.

For even $$$n$$$, there does not exist a valid array $$$a$$$ that satisfies $$$a_1=a_2=\dots=a_{n-1}=l$$$.

For even $$$n$$$, try to satisfy $$$\oplus_{i=1}^{n}a_i=0$$$.

#include<iostream>
using namespace std;
#define int long long
int t,n,l,r,k;
signed main()
{
	cin>>t;
	while(t--)
	{
		cin>>n>>l>>r>>k;
		if(n%2==1)
		{
			cout<<l<<endl;
			continue;
		}
		if(n==2)
		{
			cout<<-1<<endl;
			continue;
		}
		int res=1;
		bool fl=0;
		while(res<=r)
		{
			if(res>l)
			{
				fl=1;
				if(k<=n-2)
				 cout<<l<<endl;
				else
				 cout<<res<<endl;
				break;
			}
			res*=2;
		}
		if(!fl)
		 cout<<-1<<endl;
	}
}

#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
using namespace std;
const int N = 5005;
int t, n, mod, f[N][N], ans;
inline void add(int& x, int y){x += y;if (x >= mod) x -= mod;}
int main(){scanf("%d", &t);
	while (t --){scanf("%d%d", &n, &mod), f[0][0] = 1, ans = 0;
		for (int i = 1;i <= n;i ++) for (int j = 0;j <= i;j ++) f[i][j] = 0;
		for (int i = 1;i <= n;i ++)
		for (int j = 0, now;j < i;j ++)
		if (now = f[i - 1][j]) add(f[i][j + 1], now),
		f[i][j] = (f[i][j] + (n - i + 1ll) * (j + 1) * now) % mod;
		for (int j = 0;j <= n;j ++) add(ans, f[n][j]);printf("%d\n", ans);
	}
}