Master of GCD 线段树区间更新

题目描述
Hakase has n numbers in a line. At fi rst, they are all equal to 1. Besides, Hakase is interested in primes. She will choose a continuous subsequence [l, r] and a prime parameter x each time and for every l≤i≤r, she will change ai into ai*x. To simplify the problem, x will be 2 or 3. After m operations, Hakase wants to know what is the greatest common divisor of all the numbers.

输入
The first line contains an integer T (1≤T≤10) representing the number of test cases.
For each test case, the fi rst 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.

输出
For each test case, print an integer in one line, representing the greatest common divisor of the sequence. Due to the answer might be very large, print the answer modulo 998244353.

样例输入
2
5 3
1 3 2
3 5 2
1 5 3
6 3
1 2 2
5 6 2
1 6 2
样例输出
6
2

提示
For the first test case, after all operations, the numbers will be [6,6,12,6,6]. So the greatest common divisor is 6.

题目大意是说在一个数列中，每次对给定区间内的数乘二或者乘三，求最后所有数的最大公因数。用到了线段树的区间更新，找到区间中最少的2和最少的3，所有乘积即为答案。

#include<stdio.h>
#include<iostream>
#include<memory.h>
using namespace std;
struct Node{int l,r;long long int num2,num3;long long int lazy2,lazy3;
};
struct Node nodes[100005<<2];
void build(int l,int r,int n)
{nodes[n].l=l;nodes[n].r=r;nodes[n].lazy2=0;nodes[n].lazy3=0;nodes[n].num2=0;nodes[n].num3=0;if(l==r) return;int mid=(l+r)>>1;build(l,mid,n<<1);build(mid+1,r,n<<1|1);
}
void pushdown(int n)
{if(nodes[n].lazy2){nodes[n<<1].lazy2+=nodes[n].lazy2;nodes[n<<1|1].lazy2+=nodes[n].lazy2;nodes[n<<1].num2+=nodes[n].lazy2;nodes[n<<1|1].num2+=nodes[n].lazy2;nodes[n].lazy2=0;}if(nodes[n].lazy3){nodes[n<<1].lazy3+=nodes[n].lazy3;nodes[n<<1|1].lazy3+=nodes[n].lazy3;nodes[n<<1].num3+=nodes[n].lazy3;nodes[n<<1|1].num3+=nodes[n].lazy3;nodes[n].lazy3=0;}
}
void update(int l,int r,int x,int n)
{if(l<=nodes[n].l&&r>=nodes[n].r){if(x==2){nodes[n].lazy2++;nodes[n].num2++;}if(x==3){nodes[n].lazy3++;nodes[n].num3++;}return;}pushdown(n);int mid=(nodes[n].l+nodes[n].r)>>1;if(l<=mid) update(l,r,x,n<<1);if(r>mid) update(l,r,x,n<<1|1);nodes[n].num2=min(nodes[n<<1].num2,nodes[n<<1|1].num2);nodes[n].num3=min(nodes[n<<1].num3,nodes[n<<1|1].num3);
}
int main()
{int t;scanf("%d",&t);while(t--){int n,m;scanf("%d%d",&n,&m);build(1,n,1);for(int i=0;i<m;i++){int l,r,x;scanf("%d%d%d",&l,&r,&x);update(l,r,x,1);} long long int ans=1;for(long long int i=0;i<nodes[1].num2;i++)ans=ans*2%998244353;for(long long int i=0;i<nodes[1].num3;i++)ans=ans*3%998244353;printf("%lld\n",ans);} return 0;
}

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