题解:
$$ ans=F\left ( \prod _{i=l}^{r}a_i \right ) $$
$$ =(p_i-1){p_i}^{k_i-1}*.....*(p_j-1){p_j}^{k_j-1} $$
$$={p_i}^{k_i}*.....*{p_j}^{k_j}*(\frac{p_i-1}{p_i}*......*\frac{p_j-1}{p_j}) $$
因为数据范围保证$ a_i\leq 300 $ 所以在这个范围内只有62个素因子 我们可以直接每一位对应一bit 来判断区间中某个素因子是否出现 然后查询就行了
(常数写的有点大....懒得优化了
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <stack>
#include <queue>
#include <cmath>
#include <set>
#include <map>
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define link(x) for(edge *j=h[x];j;j=j->next)
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,r,l) for(int i=r;i>=l;i--)
const int MAXN=4e5+10;
const double eps=1e-8;
#define ll long long
const int mod=1e9+7;
using namespace std;
struct edge{int t,v;edge*next;}e[MAXN<<1],*h[MAXN],*o=e;
void add(int x,int y,int vul){o->t=y;o->v=vul;o->next=h[x];h[x]=o++;}
ll read(){ll x=0,f=1;char ch=getchar();while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}while(isdigit(ch))x=x*10+ch-'0',ch=getchar();return x*f;
}ll ksm(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod;b=b>>1;}return ans;
}ll key[MAXN<<2],tag[MAXN<<2],sum[MAXN<<2],flag[MAXN<<2];
int a[MAXN];
int vis[305],p[MAXN];void push(int rt,int l,int r){if(tag[rt]!=1){int mid=(l+r)>>1;tag[rt<<1]*=tag[rt];tag[rt<<1]%=mod;tag[rt<<1|1]*=tag[rt];tag[rt<<1|1]%=mod;sum[rt<<1]*=ksm(tag[rt],mid-l+1);sum[rt<<1]%=mod;sum[rt<<1|1]*=ksm(tag[rt],r-mid);sum[rt<<1|1]%=mod;tag[rt]=1;}if(flag[rt]){key[rt<<1]|=flag[rt];key[rt<<1|1]|=flag[rt];flag[rt<<1]|=flag[rt];flag[rt<<1|1]|=flag[rt];flag[rt]=0;}
}void up(int x){key[x]=(key[x<<1]|key[x<<1|1]);sum[x]=(sum[x<<1]*sum[x<<1|1])%mod;
}void built(int rt,int l,int r){tag[rt]=1;if(l==r){sum[rt]=1;for(int i=2;i*i<=a[l];i++){if(a[l]%i==0){int num=0;while(a[l]%i==0)num++,a[l]/=i;sum[rt]=sum[rt]*ksm(i,num)%mod;key[rt]|=(1LL<<vis[i]);}}if(a[l]!=1){sum[rt]=(sum[rt]*a[l])%mod;key[rt]|=(1LL<<vis[a[l]]);}return ;}int mid=(l+r)>>1;built(rt<<1,l,mid);built(rt<<1|1,mid+1,r);up(rt);
}ll v1,v2;void update(int rt,int l,int r,int ql,int qr){if(ql<=l&&r<=qr){tag[rt]*=v1;tag[rt]%=mod;flag[rt]|=v2;sum[rt]*=ksm(v1,r-l+1);key[rt]|=v2;return ;}int mid=(l+r)>>1;push(rt,l,r);if(ql<=mid)update(rt<<1,l,mid,ql,qr);if(qr>mid)update(rt<<1|1,mid+1,r,ql,qr);up(rt);
}ll ans1,ans2;
void query(int rt,int l,int r,int ql,int qr){if(ql<=l&&r<=qr){ans1*=sum[rt];ans1%=mod;ans2|=key[rt];return ;}int mid=(l+r)>>1;push(rt,l,r);if(ql<=mid)query(rt<<1,l,mid,ql,qr);if(qr>mid)query(rt<<1|1,mid+1,r,ql,qr);up(rt);
}int main(){int cnt=0;inc(i,2,300){bool flag=0;for(int j=2;j*j<=i;j++){if(i%j==0){flag=1;break;}}if(!flag)vis[i]=cnt++,p[cnt-1]=i;}int n=read();int m=read();inc(i,1,n)a[i]=read();built(1,1,n);char str[11];int l,r,x;while(m--){scanf("%s",str);l=read();r=read();if(str[0]=='T'){ans1=1;ans2=0;query(1,1,n,l,r);for(int i=61;i>=0;i--){if((ans2>>i)&1)ans1*=((p[i]-1)*ksm(p[i],mod-2)%mod),ans1%=mod;}printf("%lld\n",ans1);}else{x=read();v1=1;v2=0;for(int i=2;i*i<=x;i++){if(x%i==0){int num=0;while(x%i==0)num++,x/=i;v1=v1*ksm(i,num)%mod;v2|=(1LL<<vis[i]);}}if(x!=1){v1=(v1*x)%mod;v2|=(1LL<<vis[x]);}update(1,1,n,l,r);}}return 0;
}
