思路:
题目中的number与product不想冲;
即为number与product互素;
所以,求phi(product)即可;
除一个数等同于在模的意义下乘以一个数的逆元;
代码:
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define maxn 100005 #define mod 19961993 #define ll long long struct TreeNodeType { ll l,r,mid; ll dis1,dis2; }; struct TreeNodeType tree[maxn<<2]; ll n,m,cntp; ll bit[61],ans1,ans2,pi_[80],pi[80]; bool if_p[300]; inline void in(ll &now) { char Cget=getchar();now=0; while(Cget>‘9‘||Cget<‘0‘) Cget=getchar(); while(Cget>=‘0‘&&Cget<=‘9‘) { now=now*10+Cget-‘0‘; Cget=getchar(); } } ll poww(ll pos) { pos%=mod; ll mi=mod-2,res=1; while(mi) { if(mi&1) res=(res*pos)%mod; mi>>=1,pos=(pos*pos)%mod; } return res; } void tree_build(ll now,ll l,ll r) { tree[now].l=l,tree[now].r=r; if(l==r) { tree[now].dis1=bit[2],tree[now].dis2=3; return ; } tree[now].mid=l+r>>1; tree_build(now<<1,l,tree[now].mid); tree_build(now<<1|1,tree[now].mid+1,r); tree[now].dis1=tree[now<<1].dis1|tree[now<<1|1].dis1; tree[now].dis2=tree[now<<1].dis2*tree[now<<1|1].dis2%mod; } void tree_to(ll now,ll to,ll dis1,ll dis2) { if(tree[now].l==tree[now].r) { tree[now].dis1=dis1,tree[now].dis2=dis2; return ; } if(to<=tree[now].mid) tree_to(now<<1,to,dis1,dis2); else tree_to(now<<1|1,to,dis1,dis2); tree[now].dis1=tree[now<<1].dis1|tree[now<<1|1].dis1; tree[now].dis2=tree[now<<1].dis2*tree[now<<1|1].dis2%mod; } void tree_query(ll now,ll l,ll r) { if(tree[now].l==l&&tree[now].r==r) { ans1|=tree[now].dis1; ans2=(ans2*tree[now].dis2)%mod; return ; } if(l>tree[now].mid) tree_query(now<<1|1,l,r); else if(r<=tree[now].mid) tree_query(now<<1,l,r); else tree_query(now<<1,l,tree[now].mid),tree_query(now<<1|1,tree[now].mid+1,r); } int main() { for(ll i=2;i<=281;i++) { if(!if_p[i]) pi[++cntp]=i; for(ll j=1;pi[j]*i<=281&&j<=cntp;j++) { if_p[i*pi[j]]=true; if(i%pi[j]==0) break; } } for(ll i=1;i<=60;i++) { pi_[i]=poww(pi[i]); if(i==1) bit[i]=1; else bit[i]=bit[i-1]<<1; } in(n);tree_build(1,1,maxn-5); ll op,ai,bi; for(;n--;) { in(op),in(ai),in(bi); if(op==1) { ll pos=0; for(ll i=1;i<=60;i++) if(bi&&(bi%pi[i]==0)) pos+=bit[i]; bi%=mod;tree_to(1,ai,pos,bi); } else { ll ans;ans1=0,ans2=1,tree_query(1,ai,bi),ans=ans2; for(ll i=1;i<=60;i++) if(bit[i]&ans1) ans=(ans*((pi[i]-1)*pi_[i]%mod))%mod; printf("%lld\n",ans); } } return 0; }
时间: 2024-10-05 15:03:17