[POI2011] Sej-strongbox

Main topic:

A password box, the number is 0~n-1, which has a number of passwords, the characteristics of the password: if x is the password, y is the password, (x can be equal to Y) then (x+y)%n is also the password.

Give a n (<=10^14), a K (K<=min (250000,n)), the number of K (a[k]<n), the number of previous k-1 is not a password, the number of K is the password.

In 0~n-1, how many numbers are passwords?

Exercises

Recommendation (But the proof of conclusion two is not quite complete)

Seems to have some relationship with number theory.
And it must be a question of nature.

Conclusion 1: If x is the password, then gcd (n,x) is the password

found that x is the password, then k*x%n is the password.

So, there must be a t,c that makes T*X-N*C=GCD (n,x)

And according to the Pei genus theorem, it is not possible to make a smaller password with x than GCD (n,x),

Conclusion 2: If x, Y is a password, gcd (x, y) is the password.

According to the Pei genus theorem, p*x+q*y=gcd (x, y) has an integer solution.

If q is a negative q=-q, then p*x+ (c*n-q) *y=gcd (x, y) +c*n*y

Then there is a non-negative p,q that makes P*X+Q*Y=GCD (x, y) mod n

Conclusion 3: If x is the smallest of all passwords, then all passwords are x,2x,3x,... kx, and X is an approximate number of n.

Contrary Set X is the smallest, y is another password, if X is not an approximate number of Y, then gcd (x, y) <x, according to conclusion two, then gcd (x, y) is a smaller password. Contradiction.

Therefore, any y is a multiple of x.

Because for a password Z, according to the conclusion 1,GCD (N,Z) is also the password, so the minimum password x is gcd (n,z) of the approximate, that is, the approximate number of N.

So, if we find the X that satisfies the condition, then n/x is the answer.

We have the largest number of passwords, so x must be minimal.

Given a a[k] is the password, and X is an approximate number of n, so x must be GCD (a[k],n) of the approximate.

And, X can not be a[1~k-1], as long as it is, then X will be able to gather out AI, and AI is not a password contradiction.

As long as there is no such AI, then x must be the smallest password (Pei's theorem can be proved).

So we can enumerate the approximate gcd (a[k],n) and then sort it.

From small to large enumeration x, and then violence to verify whether it is a[i], the first x,n/x to match is the answer.

Code:

#include <bits/stdc++.h>using namespaceStd;typedefLong Longll;Const intn=250000+5; ll N,k;ll gcd (ll a,ll b) {returnB?GCD (b,a%b): A;} ll A[n],fac[n];inttot;intMain () {scanf ("%lld%lld",&n,&k);  for(intI=1; i<=k;i++) scanf ("%lld",&A[i]); ll G=gcd (a[k],n); //cout<< "GG" <<g<<endl;     for(LL i=1; i*i<=g;i++){        if(g%i==0) {fac[++tot]=i; if(i!=g/i) fac[++tot]=g/i; }}sort (FAC+1, fac+tot+1);  for(intI=1; i<=tot;i++){        //cout<<fac[i]<< "";        BOOLFl=true;  for(intj=1; j<=k-1; j + +){            if(a[j]%fac[i]==0) {FL=false; Break; }        }        if(fl) {printf ("%lld", N/fac[i]);return 0; }    }    //cout<< "Over" <<endl;    return 0;}

[POI2011] Sej-strongbox