$$$a \oplus 1$$$ 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 make $$$a_1=a_2=\dots=a_{n-2}=l$$$.

#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;
	}
}

Calculating the weight of a single valid sequence is difficult. Try another way?

#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);
	}
}

As same as the title, can you find a simple necessary constraint?

Consider $$$a_i = 0$$$ first.

#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
struct stu{
	int x;
	ll dp;
};
int n,a[100007],b[100007];
vector<stu> f,g;
const ll inf=0x3f3f3f3f3f3f3f3f;
void Subt(){
	cin>>n;
	for(int i=1;i<n;i++) cin>>a[i];
	for(int i=1;i<=n;i++) cin>>b[i];
	a[n]=0;
	f.clear();
	f.push_back((stu){0,0});
	for(int i=1;i<=n;i++){
		int x=0;
		g.clear();
		for(int j=30;j>=-1;j--){
			int y=x|a[i-1]|a[i];
			if(j!=-1) y|=(1<<j); 
			if(y>=b[i]){
				ll mn=inf;
				for(stu l:f){
					if((l.x&y)!=a[i-1]) continue;
					mn=min(mn,l.dp+y-b[i]);
				}
				if(mn<inf) g.push_back((stu){y,mn});
			}
			x|=((1<<j)&b[i]);
		}
		swap(f,g);
	}
	ll mn=inf;
	for(stu l:f) mn=min(mn,l.dp);
	if(mn<inf) cout<<mn<<endl;
	else cout<<-1<<endl;
	return;
}
signed main(){
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	int T=1;
	cin>>T;
	while(T--) Subt();
}

Try to determine the vertex where you were before the last move.

I just need to calculate the earliest time I can reach the end vertex.

What form of moving path from the starting vertex to the end vertex is considered optimal?

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<set>
#include<queue>
#include<unordered_map>
#include<map>
#define int long long
using namespace std;
int n,st,dist[1000002],w[1000001],tim[1000001],ans;
vector <int> g[1000001];
int f[1000001];
queue<int> q;
void getdis(int u,int fa)
{
	for(int v:g[u])
	{
		if(v==fa) continue;
		dist[v]=dist[u]+1;
		getdis(v,u);
	}
}
void dfs(int u,int fa,int k,bool inn,int hei,int TIM,int dis)
{
	if(hei<0)
	{
		TIM=max(TIM,(-hei+1)/2*2+k*2+dis);
	}
	if(TIM<=dist[u])
	{
		ans=max(ans,dist[u]);
	}
	else
	{
		return;
	}
	for(int v:g[u])
	{
		if(v==fa) continue;
		if(inn&&w[u]!=w[v])
		{
			dfs(v,u,min(k,tim[v]),inn,hei+w[v],TIM+1,dis+1);
		}
		else
		{
			dfs(v,u,k,0,hei+w[v],TIM+1,dis+1);
		}
	}
}
int T; 
signed main()
{
	ios::sync_with_stdio(false);
	cin>>T;
	int cnt=0;
	while(T--)
	{
		cnt++;
		cin>>n>>st;
		for(int i=1;i<=n;i++)
		{
			dist[i]=0;
			tim[i]=1000000000;
			g[i].clear();
			f[i]=-10000000000000000;
		}
		ans=0;
		for(int i=1;i<=n;i++)
		{
			cin>>w[i];
		}
		for(int i=1,u,v;i<n;i++)
		{
			cin>>u>>v;
			g[u].push_back(v);
			g[v].push_back(u);
		}
		getdis(1,-1);
		for(int i=1;i<=n;i++)
		{
			for(int v:g[i])
			{
				if(w[i]==1&&w[v]==1)
				{
					tim[i]=0;
					q.push(i);
					break;
				}
			}
		}
		while(!q.empty())
		{
			int u=q.front();
			q.pop();
			for(int v:g[u])
			{
				if(tim[v]==1000000000&&w[v]!=w[u])
				{
					tim[v]=tim[u]+1;
					q.push(v);
				}
			}
		}
		dfs(st,-1,tim[st],1,w[st],1,1);
		cout<<ans+(dist[st]&1)-1<<'\n';
	}
}