Hakase has n numbers in a line. At first, they are all equal to 1. Besides, Hakase is interested inprimes. She will choose a continuous subsequence [l, r] and a prime parameter x each time and forevery l ≤ i ≤ r, she will change aiinto ai ∗ x. To simplify the problem, x will be 2 or 3. After moperations, Hakase wants to know what is the greatest common divisor of all the numbers.InputThe first line contains an integer T (1 ≤ T ≤ 10) representing the number of test cases.For each test case, the first line contains two integers n (1 ≤ n ≤ 100000) and m (1 ≤ m ≤ 100000),where n refers to the length of the whole sequence and m means there are m operations.The following m lines, each line contains three integers li (1 ≤ li ≤ n), ri (1 ≤ ri ≤ n), xi (xi ∈ {2, 3}),which are referred above.OutputFor each test case, print an integer in one line, representing the greatest common divisor of thesequence. Due to the answer might be very large, print the answer modulo 998244353.

題意：

給定區間內乘2或乘3,求1到n的最大公約數。

只要想到是求1到n都有的2或3的個數，就好做了，用線段樹維護這兩個值即可。

#include <bits/stdc++.h>
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define ll long long int
using namespace std;
const int maxn=100010;
const int mod=998244353;

int n,m;
struct node{
    int l,r;
    ll sum2,sum3;//維護區間內2,3數量的最小值
}t[maxn<<2];
ll lazy2[maxn<<2],lazy3[maxn<<2];

void down(int rt){
    if(lazy2[rt]){
        ll tmp=lazy2[rt];
        lazy2[rt<<1]+=tmp,lazy2[rt<<1|1]+=tmp;
        t[rt<<1].sum2+=tmp,t[rt<<1|1].sum2+=tmp;
        lazy2[rt]=0;
    }
    if(lazy3[rt]){
        ll tmp=lazy3[rt];
        lazy3[rt<<1]+=tmp,lazy3[rt<<1|1]+=tmp;
        t[rt<<1].sum3+=tmp,t[rt<<1|1].sum3+=tmp;
        lazy3[rt]=0;
    }
}

void build(int rt,int l,int r){
    t[rt].l=l,t[rt].r=r;
    t[rt].sum2=0,lazy2[rt]=0;
    t[rt].sum3=0,lazy3[rt]=0;
    if(l==r){
        return ;
    }
    int mid=(l+r)>>1;
    build(lson),build(rson);
}

void update(int rt,int l,int r,int k){
    if(l<=t[rt].l&&r>=t[rt].r){
        if(k==2) lazy2[rt]++,t[rt].sum2++;
        if(k==3) lazy3[rt]++,t[rt].sum3++;
        return ;
    }
    down(rt);
    int mid=(t[rt].l+t[rt].r)>>1;
    if(l<=mid) update(rt<<1,l,r,k);
    if(r>mid)  update(rt<<1|1,l,r,k);
    t[rt].sum2=min(t[rt<<1].sum2,t[rt<<1|1].sum2);
    t[rt].sum3=min(t[rt<<1].sum3,t[rt<<1|1].sum3);
}

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        build(1,1,n);
        while(m--){
            int l,r,k;
            scanf("%d%d%d",&l,&r,&k);
            update(1,l,r,k);
        }
        ll ans=1;  //。。。忘了中間爆int
        for(ll i=0;i<t[1].sum2;i++)
            ans=ans*2%mod;
        for(ll i=0;i<t[1].sum3;i++)
            ans=ans*3%mod;
        printf("%lld\n",ans);
    }
    return 0;
}