HDU 2669----Romantic（扩展欧几里德求乘法逆元）

我觉得这个算法的难点是对递归调用的理解

具体的参见我上一篇博客

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 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 5110 4434 79

 

Sample Output

2 -3sorry7 -3

#include<stdio.h>int exgcd(__int64 a,__int64 b,__int64 &x,__int64 &y) {if(b==0){x=1;y=0;return a;//a是最大公约数 }/*int temp=x;    x=y;    y=temp-a/b*y;    return ans;*/__int64 ans=exgcd(b,a%b,y,x);y-=a/b*x;return ans;}int main(){   __int64 a,b,x,y;   while(scanf("%I64d%I64d",&a,&b)!=EOF)   {   __int64 gcd=exgcd(a,b,x,y);   if(gcd==1)   { x=(x%b+b)%b; y=(1-a*x)/b; printf("%I64d %I64d\n",x,y);   }   else printf("sorry\n");   }   return 0;}