uva 10673 Play with Floor and Ceil

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<iostream>

 2 #include<stdio.h>

 3 #include<cstring>

 4 #include<cstdlib>

 5 #include<cmath>

 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 }