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。

#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==0)
    {
        x=1;
        y=0;
        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<0) x=x+b;
        c=c/g;
        y=(c-x*a)/b;
        printf("%lld %lld\n",x,y);
    }
    return 0;
}