// 参考->青蛙的约会 exgcd解同余方程
定理证明->点我
1 /* 2 * @Promlem: 3 * @Time Limit: ms 4 * @Memory Limit: k 5 * @Author: pupil-XJ 6 * @Date: 2019-10-20 17:19:56 7 * @LastEditTime: 2019-10-20 18:52:09 8 */ 9 #include<cstdio> 10 #include<cstring> 11 #include<cmath> 12 #include<iostream> 13 #include<string> 14 #include<algorithm> 15 #include<vector> 16 #include<queue> 17 #include<stack> 18 #include<set> 19 #include<map> 20 #define rep(i, n) for(int i=0;i!=n;++i) 21 #define per(i, n) for(int i=n-1;i>=0;--i) 22 #define Rep(i, sta, n) for(int i=sta;i!=n;++i) 23 #define rep1(i, n) for(int i=1;i<=n;++i) 24 #define per1(i, n) for(int i=n;i>=1;--i) 25 #define Rep1(i, sta, n) for(int i=sta;i<=n;++i) 26 #define L k<<1 27 #define R k<<1|1 28 #define mid (tree[k].l+tree[k].r)>>1 29 using namespace std; 30 const int INF = 0x3f3f3f3f; 31 typedef long long ll; 32 33 inline ll gcd(ll x, ll y) { return (!y) ? x : gcd(y, x%y); } 34 35 inline ll exgcd(ll a, ll b, ll &x, ll &y) { 36 if(!b) { 37 x = 1; 38 y = 0; 39 return a; 40 } 41 ll ans = exgcd(b, a%b, x, y); 42 ll temp = x; 43 x = y; 44 y = temp - a/b*y; 45 return ans; 46 } 47 48 int main() { 49 ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); 50 ll x, y, m, n, l; 51 cin >> x >> y >> m >> n >> l; 52 ll a = l, b = n-m, c = x-y, d = gcd(a, b); 53 if(c % d != 0) cout << "Impossible "; 54 else { 55 exgcd(a, b, x, y); 56 cout << ((y*c/d)%(l/d)+(l/d))%(l/d) << " "; 57 } 58 return 0; 59 }