Extended Euclid HDU 1576

Test instructions (A/b)%9973, but because a is very large, we only give N (n=a%9973) (we given a must be divisible by B, and gcd (b,9973) = 1).

because: a%9973=n;

So: 9973*y+n=a:

Set: A/b=x; (divisible)

So: 9973*y+n=b*x;

So: b*x-9973*y=n; ① type

Also because: gcd (b,9973) = 1;

So there must be a X1*b+9973*y1=1;② type

② type *n=① Type

So just ask for a x1, you can get X, and because x=a/b, as long as the MoD 9973 is the answer.

Now give the extended Euclid:

For GCD (A, a, b) = d, the Euclidean-to-German division of (A, c) will eventually get (d, 0). At this point, for a =d, B = 0 into a*x + b*y = d, it is obvious that x = 1,y can be any value. We can use a = d, b = 0 To reverse the introduction of any GCD (A, b) = d satisfies the solution of a*x + B*y = d. If X0,y0 is the solution of B*x + (a%b) *y = d, then what is the solution for a*x + b*y = d?

B*x + (a%b) *y = D→b*x + (A-[a/b]*b) *y = d→a*y + b* (x-[a/b]*y) = d so a*x + B*y = D's solution x1 = y0,y1= x0-[a/b]*y0;

We can launch A*X+B*Y=GCD (A, B), in the lower, X and Y. Now all the problems are solved.

#include <stdio.h>
#include <cstring>
#include <iostream>
#define LL Long Long
using namespace Std; 　　　　
void Ex_gcd (ll a,ll b,ll &x,ll &y)//Extended Euclidean template, simple recursion, and finally b=0, so one layer at a layer up to the value of x, y
{
if (b==0)
{
x=1,y=0;
return; 　　
}
EX_GCD (b,a%b,x,y);
ll Temp=x;
X=y;
Y=temp-a/b*y;
return; 　　
}
int main ()
{
ll t,a,b,n,x,y;
cin>>t; 　　　　
while (t--)
{
Cin>>n>>b;
int ans;
EX_GCD (b,9973,x,y);
x=x*n;//x=a/b
Ans=c; CXc, so the

#include <stdio.h>
#include <cstring>
#include <iostream>
#define LL Long Long
using namespace Std;
void Ex_gcd (ll a,ll b,ll &x,ll &y)
{
if (b==0)
{
x=1,y=0;
return;
}
EX_GCD (B,a%b,x,y);
ll Temp=x;
X=y;
Y=temp-a/b*y;
return;
}
int main ()
{
ll T,a,b,n,x,y;
cin>>t;
while (t--)
{
cin>>n>>b;
int ans;
EX_GCD (B,9973,x,y);
X=x*n;
Ans= (x%9973+9973)%9973;//(A + b)% P = (a% p + b% p)% p, so (x%9973+9973)%9973=x%9973, if X is negative, x%9973 will go wrong, so Into (x%9973+9973)%9973, first make x positive again mod 9973
cout<<ans<<endl;
}
return 0;

}

cout<<ans<<endl;
}
return 0;

}

Extended Euclid HDU 1576