HDU 2669 Romantic (扩展欧几里得定理)

Romantic

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6643 Accepted Submission(s): 2772


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
The input contains multiple test cases.
Each case two nonnegative integer a,b (0<a, b<=2^31)

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

77 51

10 44

34 79

[align=left]Sample Output[/align]

2 -3

sorry

7 -3

注意：求 a*x+b*y=1 中的x和y，给定a和b。
（答案中的x不能是负数）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;

#define ll long long
ll n,m;

ll ex_gcd(ll a,ll b,ll &x,ll &y)
{
if (b==0)
{
x=1; y=0;
return a;
}
ll r=ex_gcd(b,a%b,y,x);
y-=a/b*x;
return r;
}
/*
ll ex_gcd(ll a,ll b,ll &x,ll &y)
{
if(!b)
{
x=1,y=0;
return a;
}
ll d=ex_gcd(b,a%b,x,y);
ll x1=x;
x=y;
y=x1-(a/b)*y;
return d;
}*/   //这段也是可以的
int main()
{
while(~scanf("%lld%lld",&n,&m))
{
ll x,y;
ll gcd=ex_gcd(n,m,x,y);
if (gcd!=1){ printf("sorry\n"); continue;}
else
{
while(x<0) x+=m;
y=(1-n*x)/m;
printf("%lld %lld\n",x,y);
}

}

return 0;
}