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简单的测试
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代码模板
/*
* TIME COMPLEXITY:O(logN)
* PARAMS:
* a ax+ny=b
* b
* n
*
* d=gcd(a,n),ax+ny=d,==>x.But the x is not minimum positive number.See below.
* the x satisfy ax==d(mod n).If x larger than n,then x=n+y.Thus a(n+y)==b(mod n)==>ay==b(mod n),do that x%=n.After that
* (ax+na)%n==b%n is correct.x+=n. The code:x%=n,x+=n,x%=n.
*
* Actually,ax'==b(mod n),and b is times of d.So change x'=x*(b/d).
* Now ax==b(mod n)[x=x'],but x is not minimum number in this equation.
* Then a'x=b'(mod n'),a'=a/d,b'=b/d,n'=n/d.According to the second line in comment,x=x%n'==>x=x%(n/d).
* Similarlly,i times n/d+ans[0].
* Last %n is another limit.
*/
vector <long long> line_mode_equation(long long a,long long b,long long n){
long long x,y;
long long d=extend_gcd(a,n,x,y);
vector<long long> ans;
ans.clear();
if(b%d==0){
x%=n;x+=n;x%=n;
ans.push_back((x*(b/d)%(n/d)+n/d)%(n/d));
for(long long i=1;i<d;i++)
ans.push_back((ans[0]+i*n/d)%n);
}
return ans;
}