I can't do anything but study number theory QAQ

I don't want to post the original English title

The question ：

There is a N*M Table for ,i That's ok j The element of the column is gcd(i,j)
Read in a length of k, The element size does not exceed 10^12 Sequence a[1..k], Ask if the sequence has ever appeared in a row of the table

1<=N,M<=10^12
1<=k<=10^4

Okay

First of all, obviously x=lcm(a[i])

then (y+i-1)%a[i]==0

namely y%[i]=1-n

And then it magically becomes the Chinese remainder theorem

Find out x and y If there is no solution, then , There's a lot going on

First of all, if x and y exceed n,m The scope of or <0 Obviously not

Then notice the enumeration i see gcd(x,y+i-1) Is it equal to a[i]

Okay ？ It doesn't seem to be much

Note that because the expression is directly defined, it is not actually given out , therefore n,m You can int burst

All the calculations require longlong？

I can't do anything but study number theory QAQ

Code ：

 #include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
#define ll long long
ll rd(){ll z=,mk=;  char ch=getchar();
while(ch<''||ch>''){if(ch=='-')mk=-;  ch=getchar();}
while(ch>=''&&ch<=''){z=(z<<)+(z<<)+ch-'';  ch=getchar();}
return z*mk;
}
ll n,m,o;  ll mo[],a[];
ll exgcd(ll a,ll b,ll &x,ll &y){
if(!b){  x=,y=;  return a;}
ll d=exgcd(b,a%b,x,y);
ll tmp=x;  x=y,y=tmp-a/b*y;
return d;
}
ll chn(){
ll M=mo[],A=a[],k,y;
for(int i=;i<=o;++i){
ll tmp=a[i]-A,d=exgcd(M,mo[i],k,y);
if(tmp%d)  return -;
ll tm=mo[i]/d;
k=(k*tmp/d%tm+tm)%tm,A+=k*M,M=M*mo[i]/d,A=(A+M)%M;
}
return A;
}
int main(){freopen("ddd.in","r",stdin);
cin>>n>>m>>o;
for(int i=;i<=o;++i)  mo[i]=rd(),a[i]=-i;
int y=chn();
ll x=,d;
for(int i=;i<=o;++i){
d=exgcd(x,mo[i],d,d);
x=x*mo[i]/d;
}
if(!y)  y=x;
if(y< || y+o->m || x>n){  cout<<"NO"<<endl;  return ;}
for(int i=;i<=o;++i)if(exgcd(x,y+i-,d,d)!=mo[i]){  cout<<"NO"<<endl;  return ;}
cout<<"YES"<<endl;
return ;
}

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