Egyptian Collegiate Programming Contest 2017 (ACM ECPC 2017) - original tests edition

题目链接：https://codeforces.com/gym/101856

D. Dream Team

题意：n个点，让你连边成为一棵树，边权为顶点的GCD（u，v）。求所有边权和的最大值。

思路：将每个数进行因数分解，从大的因子开始求最大生成树。

 #include<bits/stdc++.h>
 using namespace std;
 const int maxn = 1e5 + ;
 vector<int> G[maxn];
 int far[maxn];
 int n;
 int find(int x)
 {
     if(far[x] == x) return x;
     else return far[x] = find(far[x]);
 }
 void unite(int x, int y)
 {
     x = find(x);
     y = find(y);
     if(x == y) return ;
     far[x] = y;
 }
 bool check(int x, int y)
 {
     return find(x) == find(y);
 }
 void init()
 {
     for(int i = ;i < maxn;i++)
     {
         far[i] = i;
         G[i].clear();
     }
 }
 long long kruskal()
 {
     long long  ans = ;
     for(int i = 1e5;i >= ;i--)
     {
         if(G[i].size() < ) continue;
         else{
             int u = G[i][];
             for(int j = ;j < G[i].size();j++)
             {
                 int v = G[i][j];
                 if(!check(v, u)){
                     ans += i;
                     unite(v, u);
                 }
             }
         }
     }
     return ans;
 }
 void gao(int x, int k)
 {
     int nn = sqrt(x);
     for(int i = ;i <= nn;i++)
     {
         if(x % i == ){
             if(i * i == x){
                 G[i].push_back(k);
             }
             else {
                 G[i].push_back(k);
                 G[x/i].push_back(k);
             }
         }
     }
 }
 int main()
 {
     freopen("dream.in","r",stdin);
     int t;
     scanf("%d",&t);
     for(int cas = ; cas <= t;cas++)
     {
         init();
         scanf("%d",&n);
         int x;
         for(int i = ;i < n;i++){
             scanf("%d",&x);
             gao(x,i);
         }
         long long ans = kruskal();
         printf("Case %d: ",cas);
         printf("%lld\n",ans);
     }
     return ;
 }

  

---

F. Forgot the Flag!

题意：给你一个凸多边形，给你俩个点在多边形内，问你从a点到多边形一条边上再到b点的最小距离以及和边界的交点。

思路：对于多边形每条边求点b的对称点，答案就是a到对称点的距离，然后求直线交。注意特判俩点重合并且在线上的时候。

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int maxn = 1e5 + ;
 const double eps = 1e-;
 double INF = 1e18;
 int sgn(double x)
 {
     if(fabs(x) < eps) return ;
     else return x <  ? - : ;
 }
 struct Point{
     double x, y;
     Point(){}
     Point(double _x,double _y)
     {
         x = _x,y = _y;
     }
     void input()
     {
         scanf("%lf %lf",&x, &y);
     }
     bool operator ==(const Point &b)const{
         return sgn(x - b.x) ==  && sgn(y - b.y) == ;
     }
     bool operator <(const Point &b)const{
         return sgn(x - b.x) ==  ? sgn(y - b.y) <  : x < b.x;
     }
     Point operator -(const Point &b)const{
         return Point(x - b.x, y - b.y);
     }
     double operator ^(const Point &b)const{
         return x*b.y-y*b.x;
     }
     double operator *(const Point &b)const{
         return x*b.x + y*b.y;
     }
     double len2(){
         return x*x + y*y;
     }
     double distance(Point p){
         return hypot(x-p.x,y-p.y);
     }
     Point operator + (const Point &b)const{
         return Point(x+b.x,y+b.y);
     }
     Point operator * (const double &k)const{
         return Point(x*k,y*k);
     }
     Point operator / (const double &k)const{
         return Point(x/k,y/k);
     }

 };
 Point p[maxn];
 struct line{
     Point s, e;
     line(){}
     line(Point _s, Point _e)
     {
         s = _s,e= _e;
     }
     Point lineprog(Point p)
     {
         return s +(((e - s) * ((e - s)*(p - s)))/((e - s).len2()));
     }
     Point symmetrypoint(Point p)
     {
         Point q = lineprog(p);
         return Point(*q.x - p.x, *q.y - p.y);
     }
     Point crosspoint(line v)
     {
         double a1 = (v.e - v.s) ^ (s - v.s);
         double a2 = (v.e - v.s) ^ (e - v.s);
         return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1));
     }
     int relation(Point p){
         int c=sgn((p-s)^(e-s));
         if(c<) return ;
         else if(c>) return ;
         else return ;
     }
 };
 int main()
 {
     int t;
     freopen("flags.in","r",stdin);
     scanf("%d",&t);
     for(int cnt = ;cnt <= t;cnt++)
     {
         int n;
         scanf("%d",&n);
         for(int i = ;i < n;i++) p[i].input();
         int q;
         Point st, ed;
         line l ;
         Point a;
         scanf("%d",&q);
         printf("Case %d:\n",cnt);
         while(q--)
         {
             st.input();
             ed.input();
             double ans = INF;
             double d;
             Point anspoint;
             for(int i = ;i < n;i++){
                 l = line(p[i],p[(i+)%n]);
                 a = l.symmetrypoint(ed);
                 d = st.distance(a);
                 if(l.relation(ed) ==  && st == ed)
                 {
                     ans = ;
                     anspoint = st;
                     break;
                 }
                 if(sgn(d - ans) < ){
                     ans = d;
                     line r = line(st, a);
                     anspoint = l.crosspoint(r);
                 }
             }
             printf("%.7f %.7f %.7f\n",ans,anspoint.x,anspoint.y);
         }
     }
     return ;
 }

---

.G. Glorious Stadium

思路：推公式。

 #include<bits/stdc++.h>
 using namespace std;
 const double pi = acos(-1.0);
 int main()
 {
     int t;
     freopen("glorious.in","r",stdin);
     scanf("%d",&t);
     for(int cas = ;cas <= t;cas++)
     {
         int n;
         double r, k;
         scanf("%d %lf %lf",&n, &r, &k);
         printf("Case %d: ",cas);
         double af = pi / k;
         double c = /(cos(af)*cos(af));
         double m = (1.0-pow(c,n))/(1.0-c);
         double ans = (tan(af)*k - pi)*r*r*m;
         printf("%.5f\n",ans);
     }
     return ;
 }

---

J. Jacked Tickets

题意：将n个东西分成m份，每一种数量 i 对应 p[i] 的损失，问损失和最小多少。

思路：题目说p[i]满足凹函数的性质，那么可以贪心知道，对于每一份要尽可能的平均。

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int maxn = 1e5 + ;
 ll a[maxn];
 int main()
 {
     freopen("jacking.in","r",stdin);
     int t;
     scanf("%d",&t);
     for(int cnt = ;cnt <= t;cnt++)
     {
         int n;
         scanf("%d", &n);
         for(int i = ;i <= n;i++)
         {
             scanf("%lld",&a[i]);
         }
         int q;
         scanf("%d", &q);
         ll x, y;
         printf("Case %d:\n", cnt);
         while(q--)
         {
             scanf("%lld %lld",&x, &y);
             if(x < y){
                 printf("impossible\n");
             }
             else
             {
                 ll num = x / y;
                 ll remain = x % y;
                 ll ans = a[num]*y - a[num]*remain;
                 ans += a[num + ]*remain;
                 printf("%lld\n",ans);
             }

         }
     }
     return ;
 }

---

K. Katryoshka

题意：给定a,b,c,由（2*a,b），（2*a,b,c）,(a,b,c) 三种组成方法去构造一个模型，问最多构建几个模型。

 #include<bits/stdc++.h>
 using namespace std;
 int main()
 {
     std::ios::sync_with_stdio(false);
     freopen("katryoshka.in","r",stdin);
     int t;
     cin >> t;
     for(int cnt = ;cnt <= t; cnt++)
     {
         int a, b, c;
         cin >> a >> b >> c;
         int k = min(a, b);
         k = min(k, c);
         int ans = k;
         a -= k;b -= k;c -= k;
         ans += min(a / , c);
         cout << "Case " << cnt << ": " << ans << endl;
     }
     return ;
 }

  

---

L. Lazy ERCD

签到题。

 #include<bits/stdc++.h>
 using namespace std;

 int main(){
     freopen("lazy.in","r",stdin);
     int T;
     cin >>T;
     for(int co = ; co <= T; co++){
         int x;
         cin >>x;
         cout <<"Case "<<co<<": "<<x-<<endl;
     }
 }