[UVA][GCD] 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
網路上很多都用 double 什麼的去做, 我這裡使用 gcd 去找, 並沒有發生 TLE 的情況, 那我要怎麼保證有解, 這個我就不方便講解了。
#include <stdio.h>
#define LL long long
void gcd(LL x, LL y, LL k) {
LL t;
LL flag = 1, la = 1, lb = 0, ra = 0, rb = 1;
while(x%y) {
if(flag) {
la -= (x/y)*ra;
lb -= (x/y)*rb;
} else {
ra -= (x/y)*la;
rb -= (x/y)*lb;
}
t = x, x = y, y = t%y;
flag = 1-flag;
}
if(!flag) printf("%lld %lld\n", la*k/y, lb*k/y);
else printf("%lld %lld\n", ra*k/y, rb*k/y);
}
int main() {
LL x, k;
int t;
scanf("%d", &t);
while(t--) {
scanf("%lld %lld", &x, &k);
LL a, b;
a = x/k, b = x/k + (x%k != 0);
gcd(a, b, x);
}
return 0;
}