HDU 2669 Romantic

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

 

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

程序分析：

题中大意就是只需要求解a*x+b*y=1，a和b互质gcd(a,b)=1，这是基础情况，先想办法求解这个方程（a,b已知，x,y未知），把这个题目理解了再去求变形一点点的情况a*x+b*y=n，n是gcd(a,b)的倍数，n也可能不是1。被这个题坑死了，开始没有想到欧几里德算法，虽然案例是过了，但是一提交就是WA。之后参考了别人的代码才知道原来方法都用错了。。。

程度代码：

#include<iostream>   
#include<cstdio>   
using namespace std;  
int ext_gcd(int a,int b,int &x,int &y){  
    if(b==0){  
        x=1;y=0; return a;  
    }  
    int d=ext_gcd(b,a%b,x,y);  
    int xt=x;  
    x=y;  
    y=xt-a/b*y;  
    return d;  
}  
int main(){  
    int a,b;  
    while(scanf("%d%d",&a,&b)==2){  
        int x,y;  
        int d=ext_gcd(a,b,x,y);  
        if(d!=1)  
            printf("sorry
");  
        else{  
            int tx;  
            tx=x;  
            x=(x%b+b)%b;  
            y=y-(x-tx)/b*a;  
            printf("%d %d
",x,y);  
        }  
    }  
    return 0;  
}