HDOJ-----2669---Romantic擴展歐幾里得

Romantic

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5444    Accepted Submission(s): 2295

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 51 10 44 34 79

 

Sample Output

2 -3 sorry 7 -3

X*a + Y*b = 1，給出a， b，問有沒有x大於0的整數解，由擴展歐幾里得，知gcd(a, b) = 1有整數解，注意解得的x可能小於0，要加到大於0爲止

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<set> 
#include<queue>
#include<vector>
#include<functional>
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CL(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long LL;
const int maxn = 1e5+10;
const int MOD = 1e9+7; 
int n[20], m[20];
void ex_gcd(LL a, LL b, LL& d, LL& x, LL& y){
	if(!b){
		d = a;
		x = 1;
		y = 0;
	}
	else{
		ex_gcd(b, a%b, d, y, x);
		y -= x*(a/b);
	}
}
int main(){
	LL t, n, a, b, c, d, A, B, C, k, x, y;
	int kcase = 1;
	while(scanf("%lld%lld", &a, &b) == 2){
		ex_gcd(a, b, d, x, y);
		while(x < 0){
			x += b;
			y -= a;
		}
		if(d == 1) printf("%lld %lld\n", x, y);
		else printf("sorry\n");
	}
	return 0;
}