题意：胖子有一条大jb，大JB由n个小JB组成，每次操作将一个区间的小JB变成金银铜三者之一，最后取出所有区间的JB总价值

思路：和刷气球差不多意思了，简单的区间更新，无需更新到叶子节点，防止超时

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <queue>
#include <cmath>
#include <algorithm>
using namespace std;
#define M 100005
#define ls node<<1,l,m
#define rs node<<1|1,m+1,r
int n,m,tree1[M<<2],tree2[M<<2];
void pushdown(int node,int m)
{
    if(tree1[node])
    {
        tree1[node<<1]=tree1[node<<1|1]=tree1[node];
        tree2[node<<1]=(m-(m>>1))*tree1[node];
        tree2[node<<1|1]=(m>>1)*tree1[node];
        tree1[node]=0;
    }
}
void buildtree(int node,int l,int r)
{
    tree1[node]=0;
    tree2[node]=1;
    if(l==r) return ;
    int m=(l+r)>>1;
    buildtree(ls);
    buildtree(rs);
    tree2[node]=tree2[node<<1]+tree2[node<<1|1];
}
void update(int L,int R,int c,int node,int l,int r)
{
    if(L<=l&&r<=R)
    {
        tree1[node]=c;
        tree2[node]=c*(r-l+1);
        return ;
    }
    pushdown(node,r-l+1);
    int m=(l+r)>>1;
    if(L<=m) update(L,R,c,ls);
    if(R>m) update(L,R,c,rs);
    tree2[node]=tree2[node<<1]+tree2[node<<1|1];
}
int main()
{
    //freopen("in.txt","r",stdin);
    int t,cas=1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        buildtree(1,1,n);
        while(m--)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            update(a,b,c,1,1,n);
        }
        printf("Case %d: The total value of the hook is %d.\n",cas++,tree2[1]);
    }
    return 0;
}