hdu 赤裸欧几里得

最新推荐文章于 2018-05-30 20:39:51 发布

最新推荐文章于 2018-05-30 20:39:51 发布 · 155 阅读

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本文介绍了一种算法，用于找到满足特定线性方程Xa+Yb=1的非负整数X和整数Y，其中a和b是非负整数输入。通过扩展欧几里得算法实现，文章提供了C语言实现代码，并讨论了边界条件处理。

Now tell you two nonnegative integer a and b. Find the nonnegative integer X and integer Y to satisfy X*a + Y*b = 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 
  

Author

yifenfei

Source

HDU女生专场公开赛——谁说女子不如男

http://acm.hdu.edu.cn/webcontest/contest_showproblem.php?pid=1015&ojid=0&cid=1980&hide=0

#include<stdio.h>
int x,y;
int gcd(int a,int b)
{
__int64 t,d;
if(b==0)
{
x=1;
y=0;
return a;
}
d=gcd(b,a%b);
t=x;
x=y;
y=t-(a/b)*(y);
return d;
}
int main()
{
int a,b,d,k;
while(scanf("%d %d",&a,&b)!=EOF)
{
d=gcd(a,b);
if(d!=1) printf("sorry\n");
else
{
x=x*(1/d);
k=x/(b/d);
x=x-k*(b/d);
y=y+a/d*k;
if(x<0) {x=x+b/d;y=y-a/d;}

printf("%d %d\n",x,y);
}
}
}

通解为x=x0+b/d*t

y=y-a/d*t

注意如果x-b/d 则y+

x加则y减