hdu2669Romantic (扩展欧几里德)

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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
题意：给你两个数a，b，让你找到一个非负的整数x和一个整数y，使得ax+by=1,如果有多种情况，x要取最小的。ps:图还是挺好看的(笑)
思路：这是普通的欧几里德模板题，其中要使得x最小，那么先把特解x0求出来，那么最小的x就是(x0%b+b)%b.
#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<string>#include<algorithm>using namespace std;typedef long long ll;typedef long double ldb;#define inf 99999999#define pi acos(-1.0)ll extend_gcd(ll a,ll b,ll &x,ll &y){    if(b==0){        x=1;y=0;return a;    }    ll d=extend_gcd(b,a%b,y,x);    y-=a/b*x;    return d;}ll niyuan(ll a,ll n){    ll x,y;    ll d=extend_gcd(a,n,x,y);    if(d==1) return (x%n+n)%n;    else return -1;}ll gcd(ll a,ll b){    return b ? gcd(b,a%b) : a;}int main(){    ll n,m,b,d,a,x,y;    while(scanf("%lld%lld",&a,&b)!=EOF)    {        if(gcd(a,b)!=1){            printf("sorry\n");continue;        }        d=extend_gcd(a,b,x,y);        ll x1;        x1=(x%b+b)%b;        ll t=(x1-x)/b;        printf("%lld %lld\n",x1,y-a*t);    }    return 0;}

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