1.乘法线段树

顾名思义，就是其中的区间修改为乘法，但是呢，如果只是一个乘法，把之前的加号变成*号，然后开long long即可（因为乘法的数据超大，如果不在中间mod点儿东西还能直接超出64位）

当lz下标传递的时候，我们需要考虑乘法和加法的运算的先后顺序。我们只需要对l做这样一个处理。

lazytage分为两种，分别是加法的add和乘法的mul。

mul很简单处理，pushdown时直接*父亲的就可以了，那么加法呢？

我们需要把原先的add*父亲的mul再加上父亲的add.

所以我们pushdown函数会有所改变

void pushdown(long long i)
{
    long long k1=tree[i].mul,k2=tree[i].add;
    tree[i*2].sum=(long long)(tree[i*2].sum*k1+(k2*(tree[i*2].r-tree[i*2].l+1))%m)%m;
    tree[i*2].mul=(long long)(tree[i*2].mul*k1)%m;
    tree[i*2].add=(long long)(tree[i*2].add*k1+k2)%m;
    tree[i*2+1].sum=(long long)(tree[i*2+1].sum*k1+(k2*(tree[i*2+1].r-tree[i*2+1].l+1)%m))%m;
    tree[i*2+1].mul=(long long)(tree[i*2+1].mul*k1)%m;
    tree[i*2+1].add=(long long)(tree[i*2+1].add*k1+k2)%m;
    tree[i].add=0;
    tree[i].mul=1;
    return ;
}

2.例题 

1.模版线段树2

题解：标准的乘法线段树

#include <cstdio>
#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
long long n,q,m;
long long a[100005];
long long flag;
long long ans;
struct node{
	long long l;
	long long r;
	long long sum;
	long long mul;
	long long add;
}tree[100005*4];

void build(long long i,long long l,long long r)
{
	tree[i].l=l;
	tree[i].r=r;
	tree[i].mul=1;
	if(l==r)
	{
		tree[i].sum=a[l]%m;
		return ;
	}
	int mid=l+r>>1;
	build(i*2,l,mid);
	build(i*2+1,mid+1,r);
	tree[i].sum=(tree[i*2].sum+tree[i*2+1].sum)%m;
}


void pushdown(long long i)
{
    long long k1=tree[i].mul,k2=tree[i].add;
    tree[i*2].sum=(long long)(tree[i*2].sum*k1+(k2*(tree[i*2].r-tree[i*2].l+1))%m)%m;
    tree[i*2].mul=(long long)(tree[i*2].mul*k1)%m;
    tree[i*2].add=(long long)(tree[i*2].add*k1+k2)%m;
    tree[i*2+1].sum=(long long)(tree[i*2+1].sum*k1+(k2*(tree[i*2+1].r-tree[i*2+1].l+1)%m))%m;
    tree[i*2+1].mul=(long long)(tree[i*2+1].mul*k1)%m;
    tree[i*2+1].add=(long long)(tree[i*2+1].add*k1+k2)%m;
    tree[i].add=0;
    tree[i].mul=1;
    return ;
}

 
void add(long long i,long long l,long long r,long long k)
{
    if(tree[i].r<=r&&tree[i].l>=l)//如果当前区间被完全覆盖在目标区间里，讲这个区间的sum+k*(tree[i].r-tree[i].l+1)
    {
    	tree[i].add=(long long)(tree[i].add+k)%m;
        tree[i].sum=(long long)(tree[i].sum+k*(tree[i].r-tree[i].l+1))%m;
        return ;
    }
    pushdown(i);//向下传递
    if(tree[i*2].r>=l)
        add(i*2,l,r,k);
    if(tree[i*2+1].l<=r)
        add(i*2+1,l,r,k);
    tree[i].sum=(tree[i*2].sum+tree[i*2+1].sum)%m;
    return ;
}
 
 
void mul(long long i,long long l,long long r,long long k)
{
    if(tree[i].r<=r&&tree[i].l>=l)//如果当前区间被完全覆盖在目标区间里，讲这个区间的sum+k*(tree[i].r-tree[i].l+1)
    {
    	tree[i].mul=(long long)(tree[i].mul*k)%m;
    	tree[i].add=(long long)(tree[i].add*k)%m;
        tree[i].sum=(long long)(tree[i].sum*k)%m;
        return ;
    }
    pushdown(i);//向下传递
    if(tree[i*2].r>=l)
        mul(i*2,l,r,k);
    if(tree[i*2+1].l<=r)
        mul(i*2+1,l,r,k);
    tree[i].sum=(tree[i*2].sum+tree[i*2+1].sum)%m;
    return ;
}
 
 
long long cha(long long i,long long l,long long r)
{
    if(tree[i].l>=l && tree[i].r<=r)
        return tree[i].sum;
    if(tree[i].r<l || tree[i].l>r)  
	return 0;
    pushdown(i);
    long long s=0;
    if(tree[i*2].r>=l)  
	s=(s+cha(i*2,l,r))%m;
    if(tree[i*2+1].l<=r)  
	s=(s+cha(i*2+1,l,r))%m;
    return s;
}
 
int main()
{
	cin>>n>>q>>m;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i];
	}
	build(1,1,n);
	while(q--)
	{
		cin>>flag;
		if(flag==1)
		{
			int x,y,z;
			cin>>x>>y>>z;
			mul(1,x,y,z);
		}
		else if(flag==2)
		{
			int x,y,z;
			cin>>x>>y>>z;
			add(1,x,y,z);
		}
		else
		{
			ans=0;
			int x,y;
			cin>>x>>y;
			ans=cha(1,x,y)%m;
			printf("%lld\n",ans);
		}
	}
	return 0;
}