題意：

有一個鉤子由n個小棒構成，每個小棒有三種類型，長度分別為：1，2，3；一共有m次操作，m次操作後詢問問鉤子的長度，
注意：鉤子的初始長度為n，即每個葉子節點的值為1
AC程式碼：

//裸模板題
#include <iostream>
#include <stdio.h>
#define ll long long
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#define il inline
using namespace std;
const ll MAXN
 
 = 1e5 + 10;
struct Tree {
	ll sum;
	ll lazy;
} t[MAXN<<2];
il void push_up(ll rt) { //向上更新
	t[rt].sum = t[rt << 1].sum + t[rt << 1 | 1].sum;
}
il void push_down(ll rt, ll m) {//向下更新
	if(t[rt].lazy) { //若有標記,則將標記向下移動一層
		t[rt << 1].lazy = t[rt].lazy;
		t[rt << 1 | 1].lazy = t[
 
rt].lazy;
		t[rt << 1].sum = (m - (m >> 1)) * t[rt].lazy;
		t[rt << 1 | 1].sum = (m >> 1) * t[rt].lazy;
		t[rt].lazy = 0;//取消本層標記
	}
}
il void build(ll l,ll r, ll rt) { //建樹
	t[rt].lazy = 0;
	if(l == r) {
		t[rt].sum=1;
		return;
	}
	ll mid = (l + r) >> 1;
	build(lson);
	build
 
(rson);
	push_up(rt);//向上更新
}
il void update(ll L,ll R, ll key, ll l,ll r, ll rt) { //區間更新
	if(L <= l && R >= r) {
		t[rt].sum=(r - l + 1) * key;
		t[rt].lazy= key;
		return;
	}
	push_down(rt, r - l + 1);//向下更新
	long long int mid = (l + r) >> 1;
	if(L <= mid) update(L, R, key, lson);
	if(R > mid) update(L, R, key, rson);
	push_up(rt);//向上更新
}
il ll query(ll L, ll R, ll l, ll r,ll rt) { //區間求和
	if(L <= l && R >= r) return t[rt].sum;
	push_down(rt, r - l + 1);//向下更新
	ll mid = (l + r) >> 1;
	ll ans = 0;
	if(L <= mid) ans += query(L, R, lson);
	if(R > mid) ans += query(L, R, rson);
	return ans;
}
int main() {
	ll n, m,l,r,v;
	int t;
	cin>>t;
	for(ll i=1;i<=t;i++) {
		scanf("%lld%lld", &n, &m);
		build(1, n, 1);
		while(m--) {
			scanf("%d%d%lld",&l,&r,&v);
			update(l,r,v,1,n,1);
		}
	printf("Case %lld: The total value of the hook is %lld.\n",i,query(1,n,1,n,1));
	}
	return 0;
}