HDU 2669 Romantic（裸的拓展欧几里得）

Romantic

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8536    Accepted Submission(s):
3638

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

 

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 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <vector>
 #include <string>
 #include <queue>
 #include <stack>
 #include <algorithm>  

 #define INF 0x7fffffff
 #define EPS 1e-12
 #define MOD 1000000007
 #define PI 3.141592653579798
 #define N 100000  

 using namespace std;

 typedef long long ll;

 ll e_gcd(ll a, ll b, ll &x, ll &y)
 {
     ll d = a;
     if (b != )
     {
         d = e_gcd(b, a%b, y, x);
         y -= a / b * x;
     }
     else
     {
         x = ; y = ;
     }
     return d;
 }

 ll cal(ll a, ll b, ll c)
 {
     ll x, y;
     ll gcd = e_gcd(a, b, x, y);
     if (c%gcd != ) return -;
     x *= c / gcd;
     y *= c / gcd;
     if (b < ) b = -b;
     ll ans = x % b;//最小的x，因为ax+by=c，尽量把ax分给b，多些y，少些x
     if (ans <= ) ans += b;
     return ans;
 }

 int main()
 {
     ll a, b;
     while (cin>>a>>b)
     {
         ll ans = cal(a, b, );
         if (ans == -) printf("sorry\n");
         else printf("%I64d %I64d\n", ans, ( - ans * a) / b);
     }
     return ;
 }