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

#include

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;

}