[UVa 10673]Play with Floor and Ceil

题目大意

求不定方程 $$x = p lfloor {x over k} floor + q lceil {x over k} ceil$$ 的一组整数解，$x$,$k$给出。

题解

我们发现 $lfloor {x over k} floor$ 和 $lceil {x over k} ceil$ 只有两种情况：

1. 相差$1$，显然$gcd(lfloor {x over k} floor,lceil {x over k} ceil)=1$。显然有整数解。

2. 相等，则$k|x$，所以$gcd|x$，也有整数解。

用扩欧求出方程$$gcd(lfloor {x over k} floor,lceil {x over k} ceil) = p lfloor {x over k} floor + q lceil {x over k} ceil$$一组解，

最后将答案乘上$x over gcd$即可。

#include <set>
#include <map>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define LL long long
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define sqr(x) ((x)*(x))
using namespace std;

LL t, x, k;

LL exgcd(LL a, LL b, LL &x, LL &y) {
    if (b == 0) {
    x = 1, y = 0;
    return a;
    }
    LL c = exgcd(b, a%b, x, y);
    LL t = x;
    x = y;
    y = t-y*(a/b);
    return c;
}

int main() {
    scanf("%lld", &t);
    while (t--) {
    scanf("%lld%lld", &x, &k);
    LL p, q;
    LL tmp = exgcd(x/k, x/k+(bool)(x%k), p, q);
    printf("%lld %lld
", p*x/tmp, q*x/tmp);
    }
    return 0;
}