sincerit 2669 Romantic 扩展欧几里得

2669 Romantic
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10370 Accepted Submission(s): 4427

Problem Description
The Sky is Sprite.
The Birds is Fly in the Sky.
The Wind is Wonderful.
Blew Throw the Trees
Trees are Shaking, Leaves are Falling.
Lovers Walk passing, and so are You.
…Write in English class by yifenfei

Girls are clever and bright. In HDU every girl like math. Every girl like to solve math problem!
Now tell you two nonnegative integer a and b. Find the nonnegative integer X and integer Y to satisfy Xa + Yb = 1. If no such answer print “sorry” instead.

Input
The input contains multiple test cases.
Each case two nonnegative integer a,b (0<a, b<=2^31)
Output
output nonnegative integer X and integer Y, if there are more answers than the X smaller one will be choosed. If no answer put “sorry” instead.
Sample Input
77 51
10 44
34 79
Sample Output
2 -3
sorry
7 -3

ax + by = gcd(a,b)
当gcd(a,b) == 1时 想，即为答案
当gcd(a,b) != 1时， 先当成1计算最后结果x,y同乘与gcd(a,b)
当要求x大于0时， while (x <=0 ) x += b, y -= a;

#include <iostream>
using namespace std;
typedef long long ll;
// ax + by = 1;
ll extgcd(ll a, ll b, ll &x, ll &y) {
  ll temp;
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  } else {
    temp = extgcd(b, a%b, y, x);
    y -= (a/b) * x;
  }
  return temp;
}
int main() {
  ll n, m;
  while (cin >> n >> m) {
    ll x, y;
    ll k = extgcd(n, m, x, y);
    if (k != 1) cout << "sorry\n";
    else {
      // a(x+b) + b(y-a) == ab + ax + by - ab = 1
      while (x <= 0) {
        x += m;
        y -= n;
      }
      cout << x << " " << y << "\n";
    }
  }
  return 0;
}