Master of GCD（线段树or差分）

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

复制样例数据

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.

差分

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 998244353;
const int N = 100005;
int a[N],b[N];
ll qpow(ll a,ll b){
    ll ans=1;
    while(b){
        if(b&1){
            ans=ans*a%mod;
        }
        a=a*a%mod;
        b/=2;
    }
    return ans;
}
int main()
{
    int t;
    cin>>t;
    while(t--){
        memset(a,0,sizeof a);
        memset(b,0,sizeof b);
        int n,m;
        cin>>n>>m;
        while(m--){
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            if(z==2){
                a[x]++;
                a[y+1]--;
            }
            else {
                b[x]++;
                b[y+1]--;
            }
        }
        int s1=a[1],s2=b[1];
        for(int i=2;i<=n;i++){
            a[i]+=a[i-1];
            b[i]+=b[i-1];
            s1=min(s1,a[i]);
            s2=min(s2,b[i]);
        }
        ll ans=qpow(2,s1)%mod*qpow(3,s2)%mod;
        cout<<ans<<endl;
    }
    return 0;
}

 线段树

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
#define mod 998244353
typedef long long ll;
#define N 100005
struct xx{
    int a,b;
}sum[N*4];
struct yy{
    int a,b;
}lazy[N*4];
ll gcd (ll a,ll b){
    if(b==0) return a;
    return gcd(b,a%b);
}
void build (int l,int r,int rt){
    if(l==r) {
        sum[rt].a=0;
        lazy[rt].a=0;
        sum[rt].b=0;
        lazy[rt].b=0;
        return;
    }
    int mid =(l+r)/2;
    build(l,mid,rt*2);
    build(mid+1,r,rt*2+1);
    sum[rt].a=sum[rt*2].a+sum[rt*2+1].a;
    sum[rt].b=sum[rt*2].b+sum[rt*2+1].b;
}
void updown1(int l1,int r1,int l2,int r2,int rt){
    if(lazy[rt].a!=0){
        sum[rt*2].a+=lazy[rt].a*(r1-l1+1);
        sum[rt*2+1].a+=lazy[rt].a*(r2-l2+1);
        lazy[rt*2].a+=lazy[rt].a;
        lazy[rt*2+1].a+=lazy[rt].a;
        lazy[rt].a=0;
    }
}
void updown2(int l1,int r1,int l2,int r2,int rt){
    if(lazy[rt].b!=0){
        sum[rt*2].b+=lazy[rt].b*(r1-l1+1);
        sum[rt*2+1].b+=lazy[rt].b*(r2-l2+1);
        lazy[rt*2].b+=lazy[rt].b;
        lazy[rt*2+1].b+=lazy[rt].b;
        lazy[rt].b=0;
    }
}
void update1(int l1,int r1,int x,int l,int r,int rt){
    if(l1<=l&&r1>=r){
        sum[rt].a*=(r1-l1+1);
        lazy[rt].a++;
        return ;
    }
    int mid=(l+r)/2;
    if(l1<=mid) update1(l1,r1,x,l,mid,rt*2);
    if(r1>mid) update1(l1,r1,x,mid+1,r,rt*2+1);
    sum[rt].a=sum[rt*2].a+sum[rt*2+1].a;
}
void update2(int l1,int r1,int x,int l,int r,int rt){
    if(l1<=l&&r1>=r){
        sum[rt].b*=(r1-l1+1);
        lazy[rt].b++;
        return ;
    }
    int mid=(l+r)/2;
    if(l1<=mid) update2(l1,r1,x,l,mid,rt*2);
    if(r1>mid) update2(l1,r1,x,mid+1,r,rt*2+1);
    sum[rt].b=sum[rt*2].b+sum[rt*2+1].b;
}
ll qurry1(int p,int l,int r,int rt){
    if(p==l&&p==r){
        return sum[rt].a;
    }
    int mid=(l+r)/2;
    updown1(l,mid,mid+1,r,rt);
    ll ans=0;
    if(p<=mid) ans+=qurry1(p,l,mid,rt*2);
    else if(p>mid) ans+=qurry1(p,mid+1,r,rt*2+1);
    return ans;
}
ll qurry2(int p,int l,int r,int rt){
    if(p==l&&p==r){
        return sum[rt].b;
    }
    int mid=(l+r)/2;
    updown2(l,mid,mid+1,r,rt);
    ll ans=0;
    if(p<=mid) ans+=qurry2(p,l,mid,rt*2);
    else if(p>mid) ans+=qurry2(p,mid+1,r,rt*2+1);
    return ans;
}
int main()
{
 
    int t;
    cin>>t;
    while(t--){
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++){
            sum[i].a=0;
            sum[i].b=0;
            lazy[i].a=0;
            lazy[i].b=0;
        }
        build(1,n,1);
        while(m--){
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            if(z==2)
            update1(x,y,z,1,n,1);
            else
            update2(x,y,z,1,n,1);
        }
        ll ans1=0x3f3f3f3f,ans2=0x3f3f3f3f;
        for(int i=1;i<=n;i++){
            ll t1=qurry1(i,1,n,1);
            ans1=min(t1,ans1);
            ll t2=qurry2(i,1,n,1);
            ans2=min(ans2,t2);
           // cout<<t1<<" "<<t2<<endl;
        }
        ll ans=1;
        while(ans1--){
            ans*=2;
            ans%=mod;
        }
        while(ans2--){
            ans*=3;
            ans%=mod;
        }
        cout<<ans<<endl;
    }
    return 0;
}