sincerit 2669 Romantic 扩展欧几里得

最新推荐文章于 2019-01-06 15:36:17 发布

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本文介绍了一种使用扩展欧几里得算法解决特定线性方程的方法，当两个整数的乘积加另一个整数等于1时，如何找到满足条件的未知数。通过示例输入输出，展示了算法的具体实现过程。

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 x即为答案
当gcd(a,b) != 1时该题不考虑不为1的情况这个是扩展，先当成1计算最后结果x,y同乘与gcd(a,b)
gcd(a,b)当成1的时候，因为a, b这两个系数不变，那就相当于从x,y这两个结果约去了gcd(a,b)，最后要乘回去
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;
}