#include #include #include #include #include #include #include #include #include #include #include #include using namespace std; #define REP(i, n) for(int i = 0; i < n; i++) #define RREP(i, n) for(int i=(n)-1;i>=0;i--) #define FOR(i, b, e) for(int i = b; i < e; i++) #define to_bit(i) static_cast< bitset<8> >(i) #define INF (1<<28) #define int(n) int n; cin >> n; typedef long long ll; typedef unsigned long long ull; typedef vector VI; typedef vector VS; typedef pair PII; typedef pairPLL; typedef queue QI; typedef priority_queue maxpq; typedef priority_queue, greater > minpq; struct edge{int to, cost;}; void begin(){cin.tie(0); ios::sync_with_stdio(false);}; int gcd(int a, int b){if(a%b==0){return(b);}else{return(gcd(b,a%b));}}; int lcm(int m, int n){if((0 == m)||(0 == n)){return 0;} return ((m / gcd(m, n)) * n);}; unsigned long long comb(long n, long m){unsigned long long c = 1; m = (n - m < m ? n - m : m); for(long ns = n - m + 1, ms = 1; ms <= m; ns ++, ms ++){c *= ns; c /= ms;} return c;}; void cp(int a1[], int a2[], int l){REP(i, l) a2[i] = a1[i];}; void cp(string a1[], string a2[], int l){REP(i, l) a2[i] = a1[i];}; double sq(double d){return d*d;}; int sq(int i){return i*i;}; double sqdist(int x1, int y1, int x2, int y2){ double dx = x1 - x2, dy = y1 - y2; return dx*dx + dy*dy; }; bool inside(int y, int x, int h, int w){return 0 <= y && y <= (h-1) && 0 <= x && x <= (w-1);}; // 線分の交差判定 bool isCross(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4){ // 並行な場合 int m = (x2-x1)*(y4-y3)-(y2-y1)*(x4-x3); if(m == 0) return false; // 媒介変数の値が0より大きく1以下なら交差する、これは問題の状況によって変わるかも。 double r =(double)((y4-y3)*(x3-x1)-(x4-x3)*(y3-y1))/m; double s =(double)((y2-y1)*(x3-x1)-(x2-x1)*(y3-y1))/m; return (0 < r && r <= 1 && 0 < s && s <= 1); }; // 外積の計算 AB CD の内積を求める int crossProdct(int ax, int ay, int bx, int by, int cx, int cy, int dx, int dy){ int abx = bx - ax; int aby = by - ay; int cdx = dx - cx; int cdy = dy - cy; return abx * cdy - cdx * aby; }; double crossProdct(double ax, double ay, double bx, double by, double cx, double cy, double dx, double dy){ double abx = bx - ax; double aby = by - ay; double cdx = dx - cx; double cdy = dy - cy; return abx * cdy - cdx * aby; }; double innerProduct(double ax, double ay, double bx, double by, double cx, double cy, double dx, double dy){ double abx = bx - ax; double aby = by - ay; double cdx = dx - cx; double cdy = dy - cy; return abx * cdx + aby * cdy; }; // 三角形の内部判定 ABCの中にPがあるか判定 bool inTriangle(int ax, int ay, int bx, int by, int cx, int cy, int px, int py){ int c1 = crossProdct(ax, ay, bx, by, bx, by, px, py); int c2 = crossProdct(bx, by, cx, cy, cx, cy, px, py); int c3 = crossProdct(cx, cy, ax, ay, ax, ay, px, py); return (c1 > 0 && c2 > 0 && c3 > 0) || (c1 < 0 && c2 < 0 && c3 < 0); }; bool inTriangle(double ax, double ay, double bx, double by, double cx, double cy, double px, double py){ double c1 = crossProdct(ax, ay, bx, by, bx, by, px, py); double c2 = crossProdct(bx, by, cx, cy, cx, cy, px, py); double c3 = crossProdct(cx, cy, ax, ay, ax, ay, px, py); return (c1 > 0 && c2 > 0 && c3 > 0) || (c1 < 0 && c2 < 0 && c3 < 0); }; // 三角形の外心 void circumcenter(double ax, double ay, double bx, double by, double cx, double cy, double res[3]){ // AB AC を求める double abx = bx - ax; double aby = by - ay; double acx = cx - ax; double acy = cy - ay; double m = abx * acy - acx * aby; // 分母 m = 0 となるのは3点が一直線になるとき double s = (abx * acx + aby * acy - sq(acx) - sq(acy)) / m; // 媒介変数s res[0] = abx / 2 + s * aby / 2; res[1] = aby / 2 - s * abx / 2; // cout << res[0] << " " << res[1] << endl; res[2] = sqrt(sqdist(0, 0, res[0], res[1])); res[0] += ax; res[1] += ay; }; class BinaryIndexedTree{ public: int n; vector bit; BinaryIndexedTree(int _n){ n = _n; bit.resize(n+1,0); } int sum(int i){ int sum = 0; while(i > 0){ sum += bit[i]; i -= i & -i; } return sum; } void add(int i, int v){ while(i <= n){ bit[i] += v; i += i & -i; } } }; class UnionFindTree{ public: vector parent, rank; int n; std::set set; // 初期化 UnionFindTree(int _n){ n = _n; for(int i = 0; i < n; i++){ parent.resize(n); rank.resize(n); parent[i] = i; rank[i] = 0; } } // 根を求める int find(int x){ if(parent[x] == x){ return x; }else{ return parent[x] = find(parent[x]); } } // xとyの集合を結合 void unite(int x, int y){ x = find(x); y = find(y); if(x == y){ set.insert(x); return; } if(rank[x] < rank[y]){ parent[x] = y; set.erase(x); set.insert(y); }else{ parent[y] = x; set.erase(y); set.insert(x); if(rank[x] == rank[y]) rank[x]++; } } // xとyが同じ集合か bool same(int x, int y){ return find(x) == find(y); } // 集合の数を数える int count(){ return (int)set.size(); } }; // ワーシャルフロイド法 void warshallFloyd(int graph[100][100], int graph_size){ for(int mid_node = 0; mid_node < graph_size; mid_node++) for(int s_node = 0; s_node < graph_size; s_node++) for(int g_node = 0; g_node < graph_size; g_node++) if(s_node == g_node) graph[s_node][g_node] = 0; else graph[s_node][g_node] = min(graph[s_node][g_node], graph[s_node][mid_node] + graph[mid_node][g_node]); }; // d:隣接行列 n:グラフのサイズ s:始点 dist:始点からの距離が入る配列 void dijkstra(int graph[1000][1000], int node_count, int start_node, int distances[1000]){ REP(i, node_count) distances[i] = -1; distances[start_node] = 0; priority_queue, greater > dijkstra_pq; dijkstra_pq.push(PII(0, start_node)); while (!dijkstra_pq.empty()) { PII p = dijkstra_pq.top(); dijkstra_pq.pop(); int i = p.second; if(distances[i] < p.first) continue; for(int j = 0; j < node_count; j++){ if(graph[i][j] == -1) continue; if(distances[j] == -1){ distances[j] = distances[i] + graph[i][j]; dijkstra_pq.push(PII(distances[j], j)); }else if(distances[j] > distances[i] + graph[i][j]){ distances[j] = distances[i] + graph[i][j]; dijkstra_pq.push(PII(distances[j], j)); } } } }; // return とかの位置によってうまい具合にする int vi[4] = {-1, 1, 0, 0}, vj[4] = {0, 0, -1, 1}; void dfs2d(int i, int j, int r, int c){ // 終了条件とかをここに書く REP(k, 4){ int ni = i + vi[k]; int nj = j + vj[k]; if(inside(ni, nj, r, c)){ dfs2d(ni, nj, r, c); } } }; /** * start * @author yoshikyoto */ int main(int argc, const char * argv[]){ begin(); int a,b,c,d,e,f; cin >> a >> b >> c >> d >> e >> f; REP(i, 10000000){ a += b; if(a % d == c && a % f == e){ cout << a << endl; return 0; } } cout << -1 << endl; }