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HDU 2669 Romantic

2018-06-17 22:20:37来源：未知阅读 ()

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6409 Accepted Submission(s): 2667

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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裸的扩展欧几里得

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 #define lli long long int 
 7 using namespace std;
 8 void read(lli &n)
 9 {
10     char c='+';lli x=0;bool flag=0;
11     while(c<'0'||c>'9')
12     {c=getchar();if(c=='-')flag=1;}
13     while(c>='0'&&c<='9')
14     {x=x*10+(c-48);c=getchar();}
15     flag==1?n=-x:n=x;
16 }
17 lli x,y;
18 lli exgcd(lli a,lli b,lli & x,lli & y)
19 {
20     if(b==0)
21     {
22         x=1;y=0;
23         return a;
24     }
25     lli r=exgcd(b,a%b,y,x);
26     y-=a/b*x;
27     return r;
28 }
29 int main()
30 {
31     lli a,b;
32     while(scanf("%lld%lld",&a,&b)==2)
33     {
34         lli d=exgcd(a,b,x,y);
35         if(1%d!=0)
36             printf("sorry\n");
37         else if(d==-1)
38             printf("sorry\n");
39           else 
40         {
41             x*=1/d;
42               b=fabs(b);
43               b=b/d;
44             lli ans=x%b;
45             if(ans<=0)
46                 ans+=b;
47             printf("%lld %lld\n",ans,(1-a*ans)/b);        
48         }
49     }
50     
51     return 0;
52 }