HDU 1576 A/b (extended Euclidean algorithm)

Connection: http://acm.hdu.edu.cn/showproblem.php?pid=1576

Sample Input

21000 5387 123456789

Sample Output

79226060 Test Instructions: requirements (A/b)%9973, but because a is large, we only give N (n=a%9973) (we given a must be divisible by B, and gcd (b,9973) = 1). 

The solution of number theory: the mathematical formula pushes a push, cycle comes out, did not think of AH.

Set A = k * 9973 + N, A/b = c, c = P * 9973 + x,x is the answer we ask for. Easy to know, A = k* 9973 + N =b * P * 9973 + B * x, simplifying after k * 9973 = b * P * 9973 + b * x-n, so (b * x-n)%9973 = 0,n value know, B's value is known, and because X The value range is 0 to 9972, so the value of the enumeration x is the answer that satisfies the condition.

#include <iostream> #include <cstdio> #include <string.h>using namespace Std;int main () {    int N;    scanf ("%d", &n);    while (n--)    {        long long  n,b;        Long long X;        scanf ("%lld%lld", &n,&b);        for (int i=0;i<9973;++i)        {            if ((b*i-n)%9973 = = 0)            {                x=i;                break;            }        }        printf ("%lld\n", x);           }    return 0;}

extended Euclidean algorithm (template):

__int64 extended_euclid (__int64 a,__int64 b,__int64& x,__int64& Y) {    if (b==0)    {        x=1;        y=0;        return A;    }    __int64 r=extended_euclid (b,a%b,x,y)    __int64 temp=x;x=y;y=t-a/b*y;    return R; R is a, B, greatest common divisor}

Extended Euclidean algorithm:
The extended Euclidean algorithm is used in known non-negative integers ( Otherwise, the equation needs to be deformed, as the solution of 5x-13y=1 is deformed to 5x+ ( -13y) = 1, and then the result is processed . A, b solves a set of x, y so that Ax+by = GCD (A, B) =d (the solution must exist, according to the correlation theorem in number theory). extended Euclidean is commonly used in solving linear equations and equations.

Here is a useC + +Implementation of:
int exgcd (int a, int b, int &x, int &y)
{
if (b = = 0)
{
x = 1;
y = 0;
return A; It is difficult for---to find a value that is so realized, since the expansion of Euclid has a greater use; the individual thinks it is better to define a global array without return R.
}
int r = EXGCD (b, a% B, x, y);
int t = x;
x = y;
y = t-a/b * y;
return R;
}
ways to solve indefinite equations using the extended Euclidean algorithm:
　　 for the indefinite integer equation pa+qb=c, if C mod Gcd (A, B) = 0, the equation has an integer solution, otherwise there is no integer solution.
　　 the method for finding an integer solution is listed above, and after finding a set of solutions for P * A+q * b = GCD (A, B) p0,q0, , * P * A+q * b = GCD (A, b) other integer solution satisfies:
p = P0 +  b/gcd (A, b) * t
Q = Q0 -  a/gcd (A, b) * t (where T is an arbitrary integer, and T in P,q)
As for the integer solution of Pa+qb=c, just p * a +q * b = GCD (A, b) for each solution on the C/GCD (A, B) can be
after finding p * a+q * b = GCD (A, b) of a set of solutions p0,q0, should be
get p * a+q * b = c a set of solutions P1 = p0* (c /GCD (A, b)), q1 = q0* (b), p * A+q * b = Other integer solutions of C are satisfied:
P = P1 +  b /GCD (A, b) * t
Q = Q1 -  a/gcd (A, b) * t (where T is an arbitrary integer and T in P,q is the same )
P, q is all integer solutions for p * A+q * b = c.

Note:your Choice is C + + ide#include <iostream>using namespace std; #define K 9973int Uex (int a,int b,int &x,in T &y) {    int r;    int t;    if (b==0)    {        x=1;        y=0;        return A;    }    R=uex (b,a%b,x,y);    t=x;    x=y;    Y=t-a/b*y;    return r;} int main () {    int t,t,n,b,x,y;    scanf ("%d", &t);    while (t--)    {        scanf ("%d%d", &n,&b);        Uex (b,k,x,y);        X*=n;        if (x<0)        {            t=-x;            T=t%k;            x=k-t;     Or  do not use the T variable direct x=k-(-X)%k, or use while (x<0) {x+=k/1;}, but not recommended because it may time out! It is best to use x= (x%k+k)%k, and if statement is not required;          }        printf ("%d\n", x%k);    }    return 0;}

I haven't read it yet.

HDU 1576 A/b (extended Euclidean algorithm)