hdu 2669 Romantic 解题心得

原题：

Description

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 51 10 44 34 79

 

Sample Output

2 -3 sorry 7 -3

 

原题去掉了无用的图文。。。。

 

分析：

很直观的纯模板的扩展GCD，直接求a*x+b*y=1就可以了，注意数的大小超过int，用__int64

 

代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
__int64 exgcd(__int64 a,__int64 b,__int64 &x,__int64 &y)
{
    if(b==0) {x=1;y=0;return a;}
    else
    {
        __int64 d=exgcd(b,a%b,x,y);
        __int64 t=x;
        x=y;
        y=t-a/b*y;
        return d;
    }
}
int main()
{
    __int64 a,b,x,y;
    while(scanf("%I64d%I64d",&a,&b)!=EOF)
    {
        __int64 d=exgcd(a,b,x,y);
        if(1%d) puts("sorry");
        else
        {
             while(x<0) x+=b,y-=a;
             printf("%I64d %I64d
",x,y);
        }
    }
    return 0;
}