HDU 2669 Romantic 扩展欧几里得

Posted 2020-09-14 yjbjingcha

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Romantic

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2958    Accepted Submission(s): 1160

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

 

Author

yifenfei

 

Source

HDU女生专场公开赛——谁说女子不如男

 

通过扩展欧几里得算法求得x，y之后。要求全部的x和y的解的话。得求得通项公式：(x+k*gx , y-k*gy)  gx= b/gcd(a,b),gy = a/gcd(a,b)互素,k为随意整数 

//15MS	228K
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
#include<vector>
#define M 10007
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
long long extended_euclidean(long long n, long long m, long long &x, long long &y)
{
    if (m == 0)
    {
        x = 1;
        y = 0;
        return n;
    }
    long long g = extended_euclidean(m, n % m, x, y);
    long long t = x - n / m * y;
    x = y;
    y = t;
    return g;
}
int main()
{
    ll a,b,x,y;
    while(scanf("%I64d%I64d",&a,&b)!=EOF)
    {
        ll d=extended_euclidean(a,b,x,y);
        if(1%d)printf("sorry\n");
        else
        {
            while(x<0){x+=b;y-=a;}
            printf("%I64d %I64d\n",x,y);
        }
    }
    return 0;
}