poj1061 - 同余方程，二元一次不定方程

以前不会解二元一次不定方程的时候不会做，现在会做了。

#include <cstdio>
#include <cassert>
using namespace std;

typedef __int64 int64;

void solveType1(int64 a, int64 b, int64 & x, int64 & y, int64 & d) {
  if (b == 0) {
    x = 1; y = 0; d = a;
    return;
  }

  int64 _x, _y, _d;
  solveType1(b, a % b, _x, _y, _d);
  x = _y;
  y = _x - _y * ((a - a % b) / b);
  d = _d;
}

bool solveType2(int64 a, int64 b, int64 c, int64 & x, int64 & y, int64 & d) {
  int64 _x, _y, _d;
  solveType1(a, b, _x, _y, _d);
  // assert(a * _x + b * _y == _d);
  if (c % _d != 0) return false;
  int64 n = c / _d;
  x = n * _x;
  y = n * _y;
  d = _d;
  return true;
}

inline int64 getSign(int64 x) { return x < 0 ? -1 : 1; }
inline int64 iabs(int64 x) { return x < 0 ? -x : x; }

bool solveType3(int64 a, int64 b, int64 c, int64 & x, int64 & y, int64 & d) {
  if (c < 0) { a = -a; b = -b;  c = -c; }
  int64 sa = getSign(a);
  int64 sb = getSign(b);
  int64 _x, _y, _d;
  if (!solveType2(a * sa, b * sb, c, _x, _y, _d)) return false;
  // assert(a * sa * _x + b * sb * _y == c);
  x = sa * _x;
  y = sb * _y;
  d = _d;
  return true;
}



int64 solve(int64 x, int64 y, int64 m, int64 n, int64 L) {

  int64 t, k, _d;
  if (!solveType3(n - m, L, x - y, t, k, _d)) return -1;
  // assert((n - m) * t + L * k == x - y);
  // printf("*** %I64d %I64d %I64d\n", t, k, _d);
  int64 dx = iabs(L / _d);
  if (t == 0) return 0;
  if (t > 0) return t % dx;
  if (t < 0) return (dx - ((-t) % dx)) % dx;

}

int main() {
  int64 x, y, m, n, L;
  scanf("%I64d %I64d %I64d %I64d %I64d", &x, &y, &m, &n, &L);
  int64 ans = solve(x, y, m, n, L);
  if (ans == -1) printf("Impossible\n");
  else printf("%I64d\n", ans);
  return 0;
}