给你一个n个初始元素都为1的序列和m个询问q。
询问格式为：l r x(x为2or3)
最后求1~n所有数的GCD
GCD：把每个数分别分解质因数，再把各数中的全部公有质因数提取出来连乘，所得的积就是这几个数的最大公约数。
#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define debug() puts("++++")
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a,b,sizeof(a))
#define sz size()
#define be begin()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define mod 998244353#define pi acos(-1.0)
#define rep(i,x,n) for(int i=(x); i<(n); i++)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 1<<30;
const int maxn = 1e5+3;
const double eps = 1e-8;
const int dx[] = {-1,1,0,0,1,1,-1,-1};
const int dy[] = {0,0,1,-1,1,-1,1,-1};
int dir[2]={-1,1};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
LL t,n,m;
LL Pow(LL a, LL b)
{
LL res=1;
while(b)
{
if(b&1)
res=(res%mod * a%mod)%mod;
a = a%mod*a%mod;
b>>=1;
}
return res%mod;
}
LL a[maxn],b[maxn];
int main()
{
scanf("%lld",&t);
while(t--)
{
ms(a,0),ms(b,0);
scanf("%lld%lld",&n,&m);
while(m--)
{
LL l,r,x;
scanf("%lld%lld%lld",&l,&r,&x);
if(x==2)
{
a[l]++,a[r+1]--; //差分标记因子含有2的数、区间加维护2or3的操作数
}
else
{
b[l]++,b[r+1]--;
}
}
LL m1=a[1],m2=b[1];
for(int i=2;i<=n;i++)
{
a[i]+=a[i-1]; //前缀和维护序列本身，而序列记录2的操作数即个数，得到具体每个数的操作数
b[i]+=b[i-1];
m1=min(m1,a[i]); //2的最小操作数
m2=min(m2,b[i]);
}
LL ans = (Pow(2,m1)%mod*Pow(3,m2)%mod)%mod;
printf("%lld\n",ans);
}
}
/*
2
5 3
1 3 2
3 5 2
1 5 3
6 3
1 2 2
5 6 2
1 6 26 2
【题意】【类型】【分析】【时间复杂度&&优化】【trick】【数据】*/
