問題描述：
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
|
用的是搜索法
#include<iostream>
#include<fstream>
#include<cmath>
#include<cstdlib>
using namespace std;
ifstream fin("C:\\data30.in");
int main()
{
int T,n,k;
int f,c;
int cnt;
bool beFind;
fin>>T;
cnt=0;
while(cnt<T)
{
fin>>n>>k;
++cnt;
f=floor((double)n/k);
c=ceil((double)n/k);
beFind=false;
for(int i=0;i<=k+1;++i)
{
for(int j=0;i*f+j*c<=n;++j)
{
if(i*f+j*c==n)
{
cout<<i<<"\t"<<j<<endl;
beFind=true;
break;
}
}
if(beFind)
break;
}
}
system("pause");
return 0;
}找了個答案，用的是數論
//學會使用兩個c語言的函數，floor取下整，ceil取上整
#include <cstdio>
#include <cmath>
void gcd(long long a,long long b,long long& d,long long& x,long long& y)
{
if(!b) { d=a; x=1; y=0; }
else { gcd(b, a%b, d, y, x); y-=x*(a/b); }
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
long long a,b,c,d,k,x,y;
scanf("%lld%lld",&c,&k);
a=floor(1.*c/k); //取下整
b=ceil(1.*c/k); //取上整
gcd(a,b,d,x,y);
x*=c/d;
y*=c/d;
printf("%lld %lld\n",x,y);
}
return 0;
}http://www.cnblogs.com/scau20110726/archive/2013/02/01/2889556.html