HDU 2669 romantic

Use an Extended Euclidean algorithm.

For the first two integers $ X_1 and Y_1 $, we can calculate $ ax_1 + by_1 = gcd (a, B) $. Let's take the next step and get the formula:

\ Begin {equation} ax_1 + by_1 = gcd (a, B) = gcd (B, A \ % B) = bx_2 + (A \ % B) Y_2 \ end {equation}

In this way, $ X_1 = Y_2 $ and $ Y_1 = X_2-(A/B) Y_2 $ are obtained. In this way, we can use recursion to calculate the Extended Euclidean.

Note that the solutions that meet the conditions are not unique. For any integer $ K $, we have $ x = x + kb $, $ Y = Y-ka $, the answer is the smallest positive integer $ x $ in the numerous solutions and the paired $ y $.

The Code is as follows:

 1 #include <cstdio> 2 #include <cstdlib> 3 #include <iostream> 4 #include <cstring> 5 using namespace std; 6 typedef long long LL; 7 void exgcd(LL a, LL b, LL &d, LL &x, LL &y)  8 { 9     if( b == 0 )10     {11         d = a;12         x = 1;13         y = 0;14     }15     else16     {17         exgcd(b, a%b, d, x, y);18         int t = x;19         x = y;20         y = t - (a/b)*x;21     }22 }23 int main(int argc, char *argv[])24 {25     LL x, y, d, a, b;26     while(cin>>a>>b)27     {28         exgcd(a, b, d, x, y);29         if( d == 1 )30         {31             while( x < 0 )32             {33                 x += b;34                 y -= a;35             }36             cout<<x<<" "<<y<<endl;37         }38         else39         {40             printf ( "sorry\n" );41         }42     }43 }

 

HDU 2669 romantic