Solved:4
Rank:122
补题:8/10
A digits 2
签到 把这个数写n遍
#include <bits/stdc++.h> using namespace std; int T; int n; int main(){ scanf("%d",&T); while(T--){ scanf("%d",&n); for(int i=1;i<=n;i++)printf("%d",n); puts(""); } return 0; }
B generator 1
题意:求矩阵快速幂 幂超级大
题解:10进制模拟矩阵快速幂
#include <bits/stdc++.h> using namespace std; typedef long long ll; ll mod; struct node { ll c[2][2]; }; node mul(node x, node y) { node res; memset(res.c, 0, sizeof(res.c)); for(int i = 0; i < 2; i++) for(int j = 0; j < 2; j++) for(int k = 0; k < 2; k++) res.c[i][j] = (res.c[i][j] + x.c[i][k] * y.c[k][j] % mod) % mod; return res; } node pow_mod(node x, ll y) { node res; res.c[0][0] = res.c[1][1] = 1LL; res.c[0][1] = res.c[1][0] = 0LL; while(y) { if(y & 1) res = mul(res, x); x = mul(x, x); y >>= 1; } return res; } char s[1000005]; int ss[1000005]; int main() { ll x0, x1, a, b; cin>>x0>>x1>>a>>b; scanf("%s", s + 1); int len = strlen(s + 1); for(int i = 1; i <= len; i++) ss[i] = s[i] - '0'; cin>>mod; node A; A.c[0][0] = a; A.c[0][1] = b; A.c[1][0] = 1LL; A.c[1][1] = 0LL; if(len == 1) { if(s[1] == '1') { printf("%lld ", x1); return 0; } } ss[len]--; for(int i = len; i >= 1; i--) { if(ss[i] < 0) { ss[i] += 10; ss[i - 1]--; } } node now; now.c[0][0] = now.c[1][1] = 1LL; now.c[0][1] = now.c[1][0] = 0LL; for(int i = 1; i <= len; i++) { now = pow_mod(now, 10); int t = ss[i]; node pp = pow_mod(A, 1LL * t); now = mul(now, pp); } ll ans = now.c[0][0] * x1 % mod + now.c[0][1] * x0 % mod; ans %= mod; printf("%lld ", ans); return 0; }
C generator 2 (BSGS)
只是大概了解了下BSGS的大概思想.. 题是学弟补的 orz (我摸鱼了
#include <bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 1000000; int T; ll x0,n,a,b,p; ll ksm(ll a,ll b){ ll res = 1; for(;b;b>>=1){ if(b & 1)res = res*a%p; a = a * a % p; } return res; } pair<int,int> node[maxn]; int pos[maxn],val[maxn]; int main(){ scanf("%d",&T); while(T--){ scanf("%lld%lld%lld%lld%lld",&n,&x0,&a,&b,&p); int Q;scanf("%d",&Q); if(a == 0){ while(Q--){ int v;scanf("%d",&v); if(v == x0) puts("0"); else if(v == b) puts("1"); else puts("-1"); } continue; } ll now = x0; int up = min(n,(ll)maxn); for(int i=0;i<up;i++){ node[i] = {now,i}; now = (now * a + b) % p; } sort(node,node+up); int m = 0; for(int i=0;i<up;i++){ val[m] = node[i].first; pos[m++] = node[i].second; while(i < up - 1 && node[i].first == node[i+1].first)i++; } int inv_a = ksm(a,p-2); int inv_b = (p-b)%p*inv_a%p; ll bb = 0, aa = 1; for(int i=0;i<maxn;i++){ aa = aa * inv_a % p; bb = (bb*a%p + b) % p; } int r = p / maxn + 1; while(Q--){ int v;scanf("%d",&v); int id = lower_bound(val,val+m,v) - val; if(id < m && val[id] == v){ printf("%d ",pos[id]);continue; } if(n < maxn){ puts("-1");continue; } bool flag = false; for(int i=1;i<=r;i++){ v = (v-bb+p)%p*aa%p; int id = lower_bound(val,val+m,v) - val; if(id < m && val[id] == v){ int res = pos[id] + i * maxn; if(res >= n)puts("-1"); else printf("%d ",res); flag = true;break; } } if(!flag)puts("-1"); } } return 0; }
E independent set 1 (状压DP)
题意:最多26个点 一共有1<<26个子集 求所有子集最大独立集的和
题解:看题解补的.. 看到26个点以为int根本开不下状压的数组 结果可以用char开.. 因为每个状态的值不会超过26.. (原来还可以这样
于是对于每一个状态 选取他最小的一个点 可能有两种状态转移过来 选了这个点的贡献 和不选这个点
为什么是最小的就可以.. 因为当前状态是1个1个点慢慢选出来的 去掉这个点是子问题
还有为什么在算这个点的时候 要把和他相邻的点都扣掉.. 我最开始觉得有的点虽然和他相邻 但是不一定在那个状态下这个点产生了贡献 所以应该不扣
扣掉的点肯定不会对最大独立集产生贡献 因为是要放入的点相邻的 然后扣掉是不扣的子问题 所以没区别了
#include <bits/stdc++.h> using namespace std; char dp[1 << 26]; int adj[30]; int main() { int n, m; scanf("%d%d", &n, &m); for(int i = 1; i <= m; i++) { int u, v; scanf("%d%d", &u, &v); adj[u] |= (1 << v); adj[v] |= (1 << u); } for(int i = 0; i < n; i++) adj[i] = (~adj[i]), adj[i] ^= (1 << i); //puts("??"); for(int i = 1; i < (1 << n); i++) { int t = __builtin_ctz(i); dp[i] = max(dp[i - (1 << t)], (char)(dp[i & adj[t]] + 1)); } int ans = 0; for(int i = 1; i < (1 << n); i++) ans += dp[i]; printf("%d ", ans); return 0; }
F maximum clique 1 (最大独立集)
题意:n个数 选一个最大的子集 使得子集的元素两两不矛盾 矛盾是指两个数的二进制至少有两位不一样
题解:不矛盾的话 显然两个数xor起来不是2的幂和0 而且我们发现如果两个数的二进制1的个数奇偶性一样的话 是一定不矛盾的
于是根据二进制1的个数 建二分图 我跑的最大独立集 = 最小割
输出匹配的话 如果在最后一次bfs中 dis为INF 且是二分图中连s点的集合的点 那么就是被割掉了
dis不为INF的点 如果是连t点的集合中的点 也是被割掉了 因为这个点是肯定跑不到t点的
map常数真的大.... 学了下出题人std判2的幂
#include <bits/stdc++.h> using namespace std; const int INF = 0x3f3f3f3f; map<int, int> mp; int n, s, t, maxflow, cnt; struct node { int to, nex, val; }E[1300000]; int head[5005]; int cur[5005]; void addedge(int x, int y, int va) { E[++cnt].to = y; E[cnt].nex = head[x]; head[x] = cnt; E[cnt].val = va; E[++cnt].to = x; E[cnt].nex = head[y]; head[y] = cnt; E[cnt].val = 0; } int dis[5005]; bool inque[5005]; int st[1000005]; bool spfa() { for(int i = 1; i <= t; i++) dis[i] = INF, inque[i] = 0, cur[i] = head[i]; int top = 0; dis[s] = 0, inque[s] = 1; st[++top] = s; while(top) { int u = st[top]; top--; inque[u] = 0; for(int i = head[u]; i; i = E[i].nex) { int v = E[i].to; if(E[i].val && dis[v] > dis[u] + 1) { dis[v] = dis[u] + 1; if(!inque[v]) { inque[v] = 1; st[++top] = v; } } } } return dis[t] != INF; } int vis; int dfs(int x, int flow) { if(x == t) { vis = 1; maxflow += flow; return flow; } int used = 0; int rflow = 0; for(int i = cur[x]; i; i = E[i].nex) { cur[x] = i; int v = E[i].to; if(E[i].val && dis[v] == dis[x] + 1) { if(rflow = dfs(v, min(flow - used, E[i].val))) { used += rflow; E[i].val -= rflow; E[i ^ 1].val += rflow; if(used == flow) break; } } } return used; } void dinic() { maxflow = 0; while(spfa()) { vis = 1; while(vis) { vis = 0; dfs(s, INF); } } } bool is_power2(int x) { return x && (x & (x - 1)) == 0; } int a[5005]; int er[5005]; int main() { scanf("%d", &n); cnt = 1; s = n + 1; t = s + 1; for(int i = 1; i <= n; i++) { scanf("%d", &a[i]); int tot = 0; int tmp = a[i]; while(tmp) { if(tmp & 1) tot++; tmp >>= 1; } if(tot & 1) addedge(s, i, 1), er[i] = 1; else addedge(i, t, 1), er[i] = 0; } for(int i = 1; i <= n; i++) { for(int j = i + 1; j <= n; j++) { if(is_power2(a[i] ^ a[j])) { if(er[i]) addedge(i, j, 1); else addedge(j, i, 1); } } } dinic(); int ans = n - maxflow; printf("%d ", ans); int cn = 0; for(int i = 1; i <= n; i++) { if(dis[i] == INF) { if(!er[i]) { cn++; if(cn != ans) printf("%d ", a[i]); else { printf("%d ", a[i]); break; } } } else { if(er[i]) { cn++; if(cn != ans) printf("%d ", a[i]); else { printf("%d ", a[i]); break; } } } } return 0; }
G subsequence 1 (DP)
题意:给s，t的数字串 求s中有多少子序列10进制下大于t
题解:dp i,j,k表示s的第i位匹配到t的第j位 k = 0表示这个子序列小于t的前j位 1表示等于 2表示大于 模拟即可
#include <bits/stdc++.h> using namespace std; typedef long long ll; const ll mod = 998244353; char s[3005]; char t[3005]; ll dp[3005][3005][3]; int main() { int T; scanf("%d", &T); while(T--) { int n, m; scanf("%d%d", &n, &m); scanf("%s", s + 1); scanf("%s", t + 1); for(int i = 1; i <= n; i++) { int t1 = s[i] - '0'; for(int j = 0; j <= i; j++) dp[i][j][0] = dp[i][j][1] = dp[i][j][2] = 0; //dp[i][0][1] = 1; //dp[i][0][1] = 1; //dp[i][0][2] = 1; for(int j = 1; j <= i; j++) { int t2 = t[j] - '0'; dp[i][j][0] = dp[i - 1][j][0]; dp[i][j][1] = dp[i - 1][j][1]; dp[i][j][2] = dp[i - 1][j][2]; if(j == 1) { if(t1 == 0) continue; if(t1 == t2) dp[i][j][1]++; else if(t1 > t2) dp[i][j][2]++; else dp[i][j][0]++; } else if(j == m + 1) { dp[i][j][2] += dp[i - 1][j - 1][0] + dp[i - 1][j - 1][1] + dp[i - 1][j - 1][2]; dp[i][j][1] = 0; dp[i][j][0] = 0; } else if(j > m + 1) { dp[i][j][2] += dp[i - 1][j - 1][2]; } else { if(t1 == t2) { dp[i][j][1] += dp[i - 1][j - 1][1]; dp[i][j][2] += dp[i - 1][j - 1][2]; dp[i][j][0] += dp[i - 1][j - 1][0]; } else if(t1 > t2) { dp[i][j][2] += dp[i - 1][j - 1][1] + dp[i - 1][j - 1][2]; dp[i][j][0] += dp[i - 1][j - 1][0]; } else { dp[i][j][0] += dp[i - 1][j - 1][1] + dp[i - 1][j - 1][0]; dp[i][j][2] += dp[i - 1][j - 1][2]; } } dp[i][j][0] %= mod; dp[i][j][1] %= mod; dp[i][j][2] %= mod; } } ll ans = 0; /* for(int i = 1; i <= n; i++) { for(int j = 1; j <= i; j++) { printf("%d %d %lld %lld %lld ", i, j, dp[i][j][0], dp[i][j][1], dp[i][j][2]); } }*/ for(int i = m; i <= n; i++) { ans = (ans + dp[n][i][2]) % mod; } printf("%lld ", ans); } return 0; }
H subsequence 2 (模拟)
题意:给一个串的许多信息 每次给两种字母 然后给出在只剩下这两种字母的情况下原串的样子 输出原串
题解:每种信息更新这一种字母前有多少字母 最后组成原串
#include <bits/stdc++.h> using namespace std; char s[5]; char t[10005]; int num[15]; int pos[15][10005]; int ans[10005]; int main() { int n, m; scanf("%d%d", &n, &m); for(int i = 1; i <= 10; i++) pos[i][0] = 0; for(int i = 1; i <= n; i++) ans[i] = 0; bool f = true; for(int i = 1; i <= m * (m - 1) / 2; i++) { scanf("%s", s + 1); int len2; scanf("%d", &len2); getchar(); if(len2 == 0) continue; scanf("%s", t + 1); int c1 = s[1] - 'a' + 1; int c2 = s[2] - 'a' + 1; int cn1 = 0, cn2 = 0; if(c1 > c2) swap(c1, c2); for(int j = 1; j <= len2; j++) { if(t[j] - 'a' + 1 == c1) { cn1++; pos[c1][cn1] += cn2; } else if(t[j] - 'a' + 1 == c2) { cn2++; pos[c2][cn2] += cn1; } else { f = false; break; } } if(pos[c1][0] == 0) pos[c1][0] = cn1; else if(pos[c1][0] != cn1) f = false; if(pos[c2][0] == 0) pos[c2][0] = cn2; else if(pos[c2][0] != cn2) f = false; } for(int i = 1; i <= m; i++) { for(int j = 1; j <= pos[i][0]; j++) { //cout << pos[i][j] << " "; if(pos[i][j] + j > n) f = false; else { if(ans[pos[i][j] + j]) f = false; else ans[pos[i][j] + j] = i; } } //puts(""); } //puts("f"); //cout << f <<endl; for(int i = 1; i <= n; i++) if(!ans[i]) f = false; if(!f) { puts("-1"); } else { for(int i = 1; i <= n; i++) { printf("%c", ans[i] - 1 + 'a'); } puts(""); } return 0; }
I three points 1 (计算几何)
队友补的 (啥东西写了400行???
#include <stdio.h> #include <math.h> #include <string.h> #include <algorithm> #include <iostream> #include <vector> using namespace std; int gcd(int a, int b){ return b == 0 ? a : gcd(b, a % b); } const double eps = 1e-8; int cmp(double x) { if (fabs(x) < eps) return 0; if (x > 0) return 1; return -1; } const double pi = acos(-1.0); inline double sqr(double x) { return x * x; } struct point { double x, y; point() {} point(double a, double b) : x(a), y(b) {} void input() { scanf("%lf%lf", &x, &y); } friend point operator + (const point &a, const point &b) { return point(a.x + b.x, a.y + b.y); } friend point operator - (const point &a, const point &b) { return point(a.x - b.x, a.y - b.y); } friend bool operator == (const point &a, const point &b) { return cmp(a.x - b.x) == 0 && cmp(a.y - b.y) == 0; } friend point operator * (const point &a, const double &b) { return point(a.x * b, a.y * b); } friend point operator * (const double &a, const point &b) { return point(a * b.x, a * b.y); } friend point operator / (const point &a, const double &b) { return point(a.x / b, a.y / b); } double norm(){ return sqrt(sqr(x) + sqr(y)); } friend bool operator < (const point a, const point b){ if(a.x == b.x) return a.y < b.y; if(a.y == b.y) return a.x < b.x; return a.x < b.x; } }; double det(const point &a, const point &b) { return a.x * b.y - a.y * b.x; } double dot(const point &a, const point &b) { return a.x * b.x + a.y * b.y; } double dist(const point &a, const point &b) { return (a - b).norm(); } point rotate_point(const point &p, double A){ double tx = p.x, ty = p.y; return point(tx * cos(A) - ty * sin(A), tx * sin(A) + ty * cos(A)); } struct line { point a, b; line() {} line(point x, point y) : a(x), b(y) {} }; line point_make_line(const point a, const point b) { return line(a, b); } double dis_point_segment(const point p, const point s, const point t) { if(cmp(dot(p - s, t - s)) < 0) return (p - s).norm(); if(cmp(dot(p - t, s - t)) < 0) return (p - t).norm(); return fabs(det(s - p, t - p) / dist(s, t)); } void PointProjLine(const point p, const point s, const point t, point &cp){ double r = dot((t - s), (p - s)) / dot(t - s, t - s); cp = s + r * (t - s); } bool PointOnSegment(point p, point s, point t){ return cmp(det(p - s, t - s)) == 0 && cmp(dot(p - s, p - t)) <= 0; } bool parallel(line a, line b){ return !cmp(det(a.a - a.b, b.a - b.b)); } bool line_make_point(line a, line b, point &res){ if(parallel(a, b)) return false; double s1 = det(a.a - b.a, b.b - b.a); double s2 = det(a.b - b.a, b.b - b.a); res = (s1 * a.b - s2 * a.a) / (s1 - s2); return true; } line move_d(line a, const double &len){ point d = a.b - a.a; d = d / d.norm(); d = rotate_point(d, pi / 2); return line(a.a + d * len, a.b + d * len); } bool SegmentSamePoints(line i, line j){ return max(i.a.x, i.b.x) >= min(j.a.x, j.b.x) && max(j.a.x, j.b.x) >= min(i.a.x, i.b.x) && max(i.a.y, i.b.y) >= min(j.a.y, j.b.y) && max(j.a.y, j.b.y) >= min(i.a.y, i.b.y) && cmp(det(j.a - i.a, i.b - i.a) * det(j.b - i.a, i.b - i.a)) <= 0 && cmp(det(i.a - j.a, j.b - j.a) * det(i.b - j.a, j.b - j.a)) <= 0; } double Sabc(point a, point b, point c){ return fabs(a.x * b.y + b.x * c.y + c.x * a.y - a.x * c.y - b.x * a.y - c.x * b.y) / 2.0; } struct section{ double l, r; section() {} section(double a, double b) { l = min(a, b); r = max(a, b); } friend bool operator < (section a, section b){ if(a.l == b.l) return a.r < b.r; return a.l < b.l; } }; bool sjs(section a, section b){ if(b < a) swap(a, b); if(a.l < b.l){ return (a.r > b.l) || (b.r < a.r); } else{ return (a.r < b.r); } } const int maxn = 1005; struct polygon { int n; point a[maxn]; polygon() {} double perimeter() { double sum = 0; a[n] = a[0]; for(int i = 0; i < n; i++){ sum += (a[i + 1] - a[i]).norm(); } return sum; } double area(){ double sum = 0; a[n] = a[0]; for(int i = 0; i < n; i++){ sum += det(a[i + 1], a[i]); } return sum / 2.0; } int Point_In(point t){ int num = 0, i, d1, d2, k; a[n] = a[0]; for(i = 0; i < n; i++){ if(PointOnSegment(t, a[i], a[i + 1])) return 2; k = cmp(det(a[i + 1] - a[i], t - a[i])); d1 = cmp(a[i].y - t.y); d2 = cmp(a[i + 1].y - t.y); if(k > 0 && d1 <= 0 && d2 > 0) num++; if(k < 0 && d2 <= 0 && d1 > 0) num--; } return num != 0; } int Border_Int_Point_Num(); int Inside_Int_Point_Num(); }; int polygon::Border_Int_Point_Num(){ int num = 0; a[n] = a[0]; for(int i = 0; i < n; i++){ num += gcd(abs(int(a[i + 1].x - a[i].x)), abs(int(a[i + 1].y - a[i].y))); } return num; } int polygon::Inside_Int_Point_Num(){ return int(area()) + 1 - Border_Int_Point_Num() / 2; } struct polygon_convex{ vector<point> P; polygon_convex(int Size = 0){ P.resize(Size); } }; bool comp_less(const point &a, const point &b){ return cmp(a.x - b.x) < 0 || cmp(a.x - b.x) == 0 && cmp(a.y - b.y) < 0; } polygon_convex convex_hull(vector<point> a){ polygon_convex res(2 * a.size() + 5); sort(a.begin(), a.end(), comp_less); a.erase(unique(a.begin(), a.end()), a.end()); int m = 0; for(int i = 0; i < a.size(); ++i){ while(m > 1 && cmp(det(res.P[m - 1] - res.P[m - 2], a[i] - res.P[m - 2])) <= 0) --m; res.P[m++] = a[i]; } int k = m; for(int i = int(a.size()) - 2; i >= 0; --i){ while(m > k && cmp(det(res.P[m - 1] - res.P[m - 2], a[i] - res.P[m - 2])) <= 0) --m; res.P[m++] = a[i]; } res.P.resize(m); if(a.size() > 1) res.P.resize(m - 1); return res; } int T; double a, b, c; double w, h; bool w_h_swap = false; bool wrong(point r){ return !(cmp(r.x - 0) >= 0 && cmp(w - r.x) >= 0 && cmp(r.y - 0) >= 0 && cmp(h - r.y) >= 0); } // #define __DEBUG int main(){ scanf("%d", &T); #ifdef __DEBUG printf("%d ", T); #endif while(T--){ scanf("%lf%lf%lf%lf%lf", &w, &h, &a, &b, &c); #ifdef __DEBUG printf("%.0f %.0f %.0f %.0f %.0f ", w, h, a, b, c); #endif double l[3] = {a, b, c}; sort(l, l + 3); w_h_swap = false; point A, B, C; double dgr; A = point(0, 0); C = point(-1, -1); if(wrong(C)){ if(l[0] <= w) B = point(l[0], 0); else B = point(w, sqrt(sqr(l[0]) - sqr(w))); if(!wrong(B)){ dgr = acos( (sqr(l[0]) + sqr(l[1]) - sqr(l[2])) / (2 * l[0] * l[1]) ); C = rotate_point(B, dgr) / B.norm() * l[1]; if(wrong(C)){ dgr = acos( (sqr(l[0]) + sqr(l[2]) - sqr(l[1])) / (2 * l[0] * l[2]) ); C = rotate_point(B, dgr) / B.norm() * l[2]; } } } if(wrong(C)){ if(l[1] <= w) B = point(l[1], 0); else B = point(w, sqrt(sqr(l[1]) - sqr(w))); if(!wrong(B)){ dgr = acos( (sqr(l[1]) + sqr(l[0]) - sqr(l[2])) / (2 * l[1] * l[0]) ); C = rotate_point(B, dgr) / B.norm() * l[0]; if(wrong(C)){ dgr = acos( (sqr(l[1]) + sqr(l[2]) - sqr(l[0])) / (2 * l[1] * l[2]) ); C = rotate_point(B, dgr) / B.norm() * l[2]; } } } if(wrong(C)){ if(l[2] <= w) B = point(l[2], 0); else B = point(w, sqrt(sqr(l[2]) - sqr(w))); if(!wrong(B)){ dgr = acos( (sqr(l[2]) + sqr(l[0]) - sqr(l[1])) / (2 * l[2] * l[0]) ); C = rotate_point(B, dgr) / B.norm() * l[0]; if(wrong(C)){ dgr = acos( (sqr(l[2]) + sqr(l[1]) - sqr(l[0])) / (2 * l[2] * l[1]) ); C = rotate_point(B, dgr) / B.norm() * l[1]; } } } if(wrong(C)){ swap(w, h); w_h_swap = true; } if(wrong(C)){ if(l[0] <= w) B = point(l[0], 0); else B = point(w, sqrt(sqr(l[0]) - sqr(w))); if(!wrong(B)){ dgr = acos( (sqr(l[0]) + sqr(l[1]) - sqr(l[2])) / (2 * l[0] * l[1]) ); C = rotate_point(B, dgr) / B.norm() * l[1]; if(wrong(C)){ dgr = acos( (sqr(l[0]) + sqr(l[2]) - sqr(l[1])) / (2 * l[0] * l[2]) ); C = rotate_point(B, dgr) / B.norm() * l[2]; } } } if(wrong(C)){ if(l[1] <= w) B = point(l[1], 0); else B = point(w, sqrt(sqr(l[1]) - sqr(w))); if(!wrong(B)){ dgr = acos( (sqr(l[1]) + sqr(l[0]) - sqr(l[2])) / (2 * l[1] * l[0]) ); C = rotate_point(B, dgr) / B.norm() * l[0]; if(wrong(C)){ dgr = acos( (sqr(l[1]) + sqr(l[2]) - sqr(l[0])) / (2 * l[1] * l[2]) ); C = rotate_point(B, dgr) / B.norm() * l[2]; } } } if(wrong(C)){ if(l[2] <= w) B = point(l[2], 0); else B = point(w, sqrt(sqr(l[2]) - sqr(w))); if(!wrong(B)){ dgr = acos( (sqr(l[2]) + sqr(l[0]) - sqr(l[1])) / (2 * l[2] * l[0]) ); C = rotate_point(B, dgr) / B.norm() * l[0]; if(wrong(C)){ dgr = acos( (sqr(l[2]) + sqr(l[1]) - sqr(l[0])) / (2 * l[2] * l[1]) ); C = rotate_point(B, dgr) / B.norm() * l[1]; } } } if(w_h_swap){ swap(A.x, A.y); swap(B.x, B.y); swap(C.x, C.y); } point ans[3] = {A, B, C}; point X, Y, Z; bool flag = false; for(int i = 0; i < 3; i++){ for(int j = 0; j < 3; j++){ if(j == i) continue; for(int k = 0; k < 3; k++){ if(k == i || k == j) continue; X = ans[i]; Y = ans[j]; Z = ans[k]; if(cmp(dist(X, Y) - a) == 0 && cmp(dist(X, Z) - b) == 0 && cmp(dist(Y, Z) - c) == 0){ flag = true; break; } } if(flag) break; } if(flag) break; } printf("%.8f %.8f %.8f %.8f %.8f %.8f ", X.x, X.y, Y.x, Y.y, Z.x, Z.y); } return 0; }