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

---

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。

 #include<iostream>
 #include<stdio.h>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 using namespace std;
 typedef long long LL;

 LL Ex_GCD(LL a,LL b,LL &x,LL& y)
 {
     if(b==)
     {
         x=;
         y=;
         return a;
     }
     LL g=Ex_GCD(b,a%b,x,y);
     LL hxl=x-(a/b)*y;
     x=y;
     y=hxl;
     return g;
 }
 int main()
 {
     LL T;
     LL a,b,c,g,x,y,n,m;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%lld %lld",&n,&m);
         a = (LL)floor(n/(double)m);/**向下取整floor(1.999) = 1  **/
         b = (LL)ceil(n/(double)m);
         c = n;
         g = Ex_GCD(a,b,x,y);

         x=x*(c/g);
         b=b/g;
         x=x%b;
         while(x<) x=x+b;
         c=c/g;
         y=(c-x*a)/b;
         printf("%lld %lld\n",x,y);
     }
     return ;
 }