刚刚学完就来敲了，敲完还和答案核对了一遍，没任何差别（可能我有的地方没看到？）

代码:

#include<bits/stdc++.h>
using namespace std;
const int M = 1e5 + 10; typedef long long ll;
int n, m, r, p, cnt, wson[M], vl[M], sz[M], fa[M], dfn[M], rdfn[M], top[M], d[M];
ll sgt[M * 4];
vector<int> tr[M];
ll read(){
	int f = 1, a = 0;
	char c = getchar();
	while(c < '0' || c > '9'){
		if (c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9'){
		a = (a << 1) + (a << 3) + (c ^ 48);
		c = getchar();
	}
	return f * a;
}
dfs1(int u, int f){//求父节点， 重儿子
	fa[u] = f;
	sz[u] = 1;
	for(int i = 0; i < tr[u].size(); ++i){
		int v = tr[u][i];
		if(v == f) continue;
		d[v] = d[u] + 1;
		dfs1(v, u);
		sz[u] += sz[v];
		if(sz[v] > sz[wson[u]]) wson[u] = v; 
	}
}
dfs2(int u, int topp){//求重链顶，dfs序
	dfn[u] = ++cnt;
	rdfn[cnt] = u;
	top[u] = topp;
	if(wson[u] != 0){
		dfs2(wson[u], topp);
		for(int i = 0; i < tr[u].size(); ++i)
			dfs2(tr[u][i], tr[u][i]);
	}
}
ll lzt[M * 4];//线段树，因为看过MLE的案例所以没写结构体，见谅
void pushup(int u){sgt[u] = sgt[u * 2] + sgt[u * 2 + 1] % p;}
void build(int u, int L, int R){
	if(L == R){
		sgt[u] = vl[rdfn[L]];
		return;
	}
	int mid = L + (R - L >> 1);
	build(u * 2, L, mid), build(u * 2 + 1, mid + 1, R);
	pushup(u);
}
void maketag(int u, int len, int vai){
	vl[u] += vai * len % p;
	lzt[u] += vai % p;
	lzt[u] %= p, vl[u] %= p;
}
void pushdown(int u, int L, int R){
	int mid = L + (R - L >> 1);
	maketag(u * 2, mid - L + 1, lzt[u]);
	maketag(u * 2 + 1, R - mid, lzt[u]);
	lzt[u] = 0;
}
void update(int u, int L, int R, int l, int r, int vai){
	if(l <= L && R <= r){
		maketag(u, R - L + 1, vai);
	}
	else if(!(r < L || R < l)){
		int mid = L + (R - L >> 1);
		pushdown(u, L, R);
		update(u * 2, L, mid, l, r, vai);
		update(u * 2 + 1, mid + 1, R, l, r, vai);
		pushup(u);
	}
}
ll query(int u, int L, int R, int l, int r){
	if(l <= L && R <= r)
		return sgt[u];
	else if(!(r < L || R < l)){
		int mid = L + (R - L >> 1);
		pushdown(u, L, R);
		return (query(u * 2, L, mid, l, r) + query(u * 2 + 1, mid + 1, R, l, r));
	}
	else return 0;
}
void upd(int u, int v, int vai){//题目描述上的1，2操作
	while(top[u] != top[v]){
		if(d[top[u]] < d[top[v]]) swap(u, v);
		update(1, 1, n, dfn[top[u]], dfn[u], vai);
		u = fa[top[u]];
	}
	update(1, 1, n, min(dfn[u], dfn[v]), max(dfn[u], dfn[v]), vai);
}
ll qry(int u, int v){
	ll re = 0;
	while(top[u] != top[v]){
		if(d[top[u]] < d[top[v]]) swap(u, v);
		re += query(1, 1, n, dfn[top[u]], dfn[u]) % p;
		re %= p;
		u = fa[top[u]];
	}
	re += query(1, 1, n, min(dfn[u], dfn[v]), max(dfn[u], dfn[v])) % p;
	return re % p;
}
int main(){
	n = read(), m = read(), r = read(), p = read();
	for(int i = 1; i <= n; ++i) vl[i] = read();
	for(int i = 1, u, v; i < n; ++i){
		u = read(), v = read();
		tr[u].push_back(v);
		tr[v].push_back(u);
	} 
	build(1, 1, n); ++d[r]; dfs1(r, 0); dfs2(r, 0);
	for(int i = 1, x, y, z, t; i <= m; ++i){
		t = read();
		if(t == 1){
			x = read(), y = read(), z = read();
			upd(x, y, z);
		}
		else if(t == 2){
			x = read(), y = read();
			cout << qry(x, y) << "\n";
		}
		else if (t == 3){
			x = read(), y = read();
			update(1, 1, n, dfn[x], dfn[x] + sz[x] - 1, y);
		}
		else{
			x = read();
			cout << query(1, 1, n, dfn[x], dfn[x] + sz[x] - 1) % p << "\n";
		}
	}
	return 0;
}