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。
1 #include <set> 2 #include <map> 3 #include <ctime> 4 #include <cmath> 5 #include <queue> 6 #include <stack> 7 #include <vector> 8 #include <cstdio> 9 #include <string> 10 #include <cstring> 11 #include <cstdlib> 12 #include <iostream> 13 #include <algorithm> 14 #define LL long long 15 #define Max(a, b) ((a) > (b) ? (a) : (b)) 16 #define Min(a, b) ((a) < (b) ? (a) : (b)) 17 #define sqr(x) ((x)*(x)) 18 using namespace std; 19 20 LL t, x, k; 21 22 LL exgcd(LL a, LL b, LL &x, LL &y) { 23 if (b == 0) { 24 x = 1, y = 0; 25 return a; 26 } 27 LL c = exgcd(b, a%b, x, y); 28 LL t = x; 29 x = y; 30 y = t-y*(a/b); 31 return c; 32 } 33 34 int main() { 35 scanf("%lld", &t); 36 while (t--) { 37 scanf("%lld%lld", &x, &k); 38 LL p, q; 39 LL tmp = exgcd(x/k, x/k+(bool)(x%k), p, q); 40 printf("%lld %lld ", p*x/tmp, q*x/tmp); 41 } 42 return 0; 43 }