題幹

思路

裴蜀定理：n元線性方程有解當且僅當s是gcd（a，b，c……，z）的倍數

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll N=1e5+100;
const ll M=1e6+100;
ll read()
{
	char ch=' ';
	ll f=1;ll x=0;
	while(ch<'0'||ch>'9')
	{
		if(ch=='-') f=-1;ch=getchar();
	}
	while(ch>='0'&&ch<='9')
	{
	    x=x*10+ch-'0';ch=getchar();
	}
	return x*f;
}
struct node
{
	ll v,nxt;
	ll w;
}edge[M];
ll head[N],cnt;
void add(ll u,ll v,ll w)
{
	cnt++;
	edge[cnt].v=v;
	edge[cnt].w=w;
	edge[cnt].nxt=head[u];
	head[u]=cnt;
}
ll dis[N];
void dfs(ll u,ll fa)
{
	for(ll i=head[u];i;i=edge[i].nxt)
	{
		ll v=edge[i].v;
		ll w=edge[i].w;
		if(v!=fa)
		{
			dis[v]=dis[u]+w;
			dfs(v,u);
		}
	}
}
ll gcd(ll a,ll b)
{
	return b==0?a:gcd(b,a%b);
}
int main()
{
	ll n,q;
	n=read();q=read();
	ll i,j;
	for(i=1;i<n;i++)
	{
		ll u,v,w;
		u=read();v=read();w=read();
		add(u,v,w);
		add(v,u,-w);
	}
	dis[1]=0;
	dfs(1,0);
	ll hh=-1;
	ll gg=0;
	for(i=1;i<=q;i++)
	{
		ll op,x,y,k;
		op=read();x=read();y=read();k=read();
		if(op==0)
		{
			ll tmp=dis[x]-dis[y]+k;
			tmp=abs(tmp);
			if(hh==-1)
			{
				hh=1;
				gg=tmp;
			}
			else
			{
				gg=gcd(gg,tmp);
			}
		}
		else if(op==1)
		{
			ll tmp=dis[y]-dis[x];
			if(hh==-1)
			{
				if(tmp==k)
				{
					puts("yes");
				}
				else
				{
					puts("no");
				}
			}
			else
			{
				if(abs(tmp-k)%gg==0) puts("yes");
				else puts("no");
			}
		}
	}
	return 0;
}