HDU 2669 Romantic 擴展歐幾裡得

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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
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女生專場公開賽——誰說女子不如男
通過擴展歐幾裡得算法求得x，y之後，要求所有的x和y的解的話，得求得通項公式：(x+k*gx , y-k*gy) gx= b/gcd(a,b),gy = a/gcd(a,b)互素,k為任意整數

//15MS	228K
#include
#include
#include
#include
#include
#include
#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;
}