HDU 2669 Romantic, hdu2669romantic

HDU 2669 Romantic, hdu2669romantic

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

Problem DescriptionThe 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.

 

InputThe input contains multiple test cases.
Each case two nonnegative integer a, B (0 <a, B <= 2 ^ 31)

 

Outputoutput 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 Input77 5110 4434 79

 

Sample Output2-3sorry7-3

 

Authoryifenfei

 

SourceHDU girls' session Open-who said women are not as good as men?

 

Recommendlcy | We have carefully selected several similar problems for you: 2668 2674 2671 2670 2672 bare Extended Euclidean

 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 }