线段树

1.

题意：输入t，为有t组数据，每组数据先输入 n，m表示有n个点，m个操作，下面有m行，每行有三个数，前两个为一个区间，把这个区间中的 每个数都乘以最后一个数，最后一个数只能是 2,3；求，最后这个n个数的最大公倍数，刚开始这n个数都是1

线段树记录区间中出现2的次数和3的次数

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
#define MAX 100050
typedef long long ll;
struct Tree
{
   int two,three;
   int lazy_two;
   int lazy_three;
   int l,r;
}tree[4*MAX];

int n,m;
void push_up(int i)
{
    tree[i].three=min(tree[i<<1].three,tree[i<<1|1].three);
    tree[i].two=min(tree[i<<1].two,tree[i<<1|1].two);
}
void push_down(int i)
{
    if(tree[i].lazy_two)
    {
        tree[i<<1].two+=tree[i].lazy_two;
        tree[i<<1].lazy_two+=tree[i].lazy_two;
        tree[i<<1|1].two+=tree[i].lazy_two;
        tree[i<<1|1].lazy_two+=tree[i].lazy_two;
        tree[i].lazy_two=0;
    }
    if(tree[i].lazy_three)
    {
        tree[i<<1].three+=tree[i].lazy_three;
        tree[i<<1].lazy_three+=tree[i].lazy_three;
        tree[i<<1|1].three+=tree[i].lazy_three;
        tree[i<<1|1].lazy_three+=tree[i].lazy_three;
        tree[i].lazy_three=0;
    }
}
void build(int l,int r,int i)
{
    tree[i].l=l;
    tree[i].r=r;
    if(l==r)
    {
            tree[i].three=0;
            tree[i].two=0;
            tree[i].lazy_three=0;
            tree[i].lazy_two=0;
            return ;
    }
    int mid=(l+r)>>1;
    build(l,mid,i<<1);
    build(mid+1,r,i<<1|1);
    push_up(i);
}

void updata(int l,int r,int v,int i)
{
    if(tree[i].l>=l&&tree[i].r<=r)
    {
        if(v==2)
        {
            tree[i].two++;
            tree[i].lazy_two++;
            return ;
        }
        if(v==3)
        {
            tree[i].three++;
            tree[i].lazy_three++;
            return ;
        }
    }
    push_down(i);
    int mid=(tree[i].l+tree[i].r)>>1;
    if(mid>=l) updata(l,r,v,i<<1);
    if(mid<r)  updata(l,r,v,i<<1|1);
    push_up(i);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
       memset(tree,0,sizeof(tree));
       scanf("%d%d",&n,&m);
       build(1,n,1);
       for(int i=0;i<m;i++)
       {
           int l,r,v;
           scanf("%d%d%d",&l,&r,&v);
           updata(l,r,v,1);
       }
       ll ans=1;
       for(int i=0;i<tree[1].two;i++)
        ans=(ans*2)%998244353;
       for(int i=0;i<tree[1].three;i++)
        ans=(ans*3)%998244353;
       //cout <<two<<endl;
       //cout <<three<<endl;
       cout <<ans<<endl;

    }

    return 0;
}