Problem A

Play with Floor and Ceil

Input: standard input

Output: standard output

Time Limit: 1 second

 

Theorem

For any two integers x and k there exists two more integers p and q such that:

 

It’s a fairly easy task to prove this theorem, so we’d not ask you to do that. We’d ask for something even easier! Given the values of x and k, you’d only need to find integers p and q that satisfies the given equation.

 

Input

The first line of the input contains an integer, T (1≤T≤1000) that gives you the number of test cases. In each of the following T lines you’d be given two positive integers x and k. You can safely assume that x and k will always be less than 10⁸.

 

Output

For each of the test cases print two integers: p and q in one line. These two integers are to be separated by a single space. If there are multiple pairs of p and q that satisfy the equation, any one would do. But to help us keep our task simple, please make sure that the values,  and fit in a 64 bit signed integer.

 

Sample Input                              Output for Sample Input

3 5 2 40 2 24444 6

1 1 1 1 0 6

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Problem setter: Monirul Hasan, Member of Elite Problemsetters' Panel

Special Thanks: Shahriar Manzoor, Member of Elite Problemsetters' Panel

 

知道floor和ceil 就行。

floor向下取整 floor(1.9999) = 1;

同理。

1 #include
     
       2 #include
      
        3 #include
       
         4 #include
        
          5 #include
         
           6 using namespace std; 7 typedef long long LL; 8  9 LL Ex_GCD(LL a,LL b,LL &x,LL& y)10 {11     if(b==0)12     {13         x=1;14         y=0;15         return a;16     }17     LL g=Ex_GCD(b,a%b,x,y);18     LL hxl=x-(a/b)*y;19     x=y;20     y=hxl;21     return g;22 }23 int main()24 {25     LL T;26     LL a,b,c,g,x,y,n,m;27     scanf("%d",&T);28     while(T--)29     {30         scanf("%lld %lld",&n,&m);31         a = (LL)floor(n/(double)m);/**向下取整floor(1.999) = 1  **/32         b = (LL)ceil(n/(double)m);33         c = n;34         g = Ex_GCD(a,b,x,y);35 36         x=x*(c/g);37         b=b/g;38         x=x%b;39         while(x<0) x=x+b;40         c=c/g;41         y=(c-x*a)/b;42         printf("%lld %lld\n",x,y);43     }44     return 0;45 }