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HDU 2669 Romantic
2017-10-12 12:51:43 】 浏览:2406
Tags:HDU 2669 Romantic

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
 
 

 

Recommend
lcy   |   We have carefully selected several similar problems for you:   2668  2674  2671  2670  2672 
 
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 }

 

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