扩展欧几里德算法求逆元3

 1 int gcd(int x3,int y3)
 2 {
 3     int x1 = 1,x2 = 0,y1 = 0,y2 = 1;
 4     while(1)
 5     {
 6         if (y3==1) return y2;
 7         int q=x3/y3;
 8         int t1=x1-q*y1,t2=x2-q*y2,t3=x3-q*y3;
 9         x1 = y1,x2 = y2,x3 = y3;
10         y1 = t1,y2 = t2,y3 = t3;
11     }
12 }
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原文地址:https://www.cnblogs.com/wsruning/p/4674226.html