2018.9.25 NOIP模拟赛

*注意：这套题目应版权方要求，不得公示题面。

从这里开始

• Problem A XOR
• Problem B GCD
• Problem C SEG

　　表示十分怀疑出题人水平，C题数据和标程都是错的。有原题，差评。

Problem A XOR

题目大意

　　最小异或生成树

　　出门左拐Codeforces 888G。

Code

 #include <iostream>
 #include <cstdlib>
 #include <cstdio>
 #include <vector>
 #ifndef WIN32
 #define Auto "%lld"
 #else
 #define Auto "%I64d"
 #endif
 using namespace std;
 typedef bool boolean;
 #define ll long long

 const int Mask =  << ;
 const signed int inf = (signed) (~0u >> );

 typedef class TrieNode {
     public:
         TrieNode* ch[];
 }TrieNode;

 TrieNode pool[];
 TrieNode* top;

 TrieNode* newnode() {
     top->ch[] = top->ch[] = NULL;
     return top++;
 }

 typedef class Trie {
     public:
         TrieNode* rt;

         void reset() {
             top = pool;
             rt = newnode();
         }

         void insert(int x) {
             TrieNode* p = rt;
             int mask = Mask;
             while (mask) {
                 int c = ((x & mask) ? () : ());
                 if (!p->ch[c])
                     p->ch[c] = newnode();
                 p = p->ch[c], mask >>= ;
             }
         }

         int query(int x) {
             TrieNode* p = rt;
             int mask = Mask, rt = ;
             while (mask) {
                 int c = ((x & mask) ? () : ());
                 if (p->ch[c])
                     p = p->ch[c];
                 else
                     p = p->ch[c ^ ], rt += mask;
                 mask >>= ;
             }
             return rt;
         }
 }Trie;

 int n;
 int* ar;
 Trie tr;

 inline void init() {
     scanf("%d", &n);
     ar = new int[(n + )];
     for (int i = ; i <= n; i++)
         scanf("%d", ar + i);
 }

 ll dividing(int tem, vector<int>& vs) {
     if (!tem)
         return ;
     vector<int> v0, v1;
     for (int i = ; i < (signed) vs.size(); i++)
         if (vs[i] & tem)
             v1.push_back(vs[i]);
         else
             v0.push_back(vs[i]);
     if (!v0.size())
         return dividing(tem >> , v1);
     if (!v1.size())
         return dividing(tem >> , v0);
     tr.reset();
     for (int i = ; i < (signed) v0.size(); i++)
         tr.insert(v0[i]);
     int rt = inf;
     for (int i = , cmp; i < (signed) v1.size(); i++) {
         cmp = tr.query(v1[i]);
         rt = min(cmp, rt);
     }
     return rt * 1ll + dividing(tem >> , v0) + dividing(tem >> , v1);
 }

 inline void solve() {
     vector<int> vs;
     for (int i = ; i <= n; i++)
         vs.push_back(ar[i]);
     printf(Auto, dividing(Mask, vs));
 }

 int main() {
     freopen("A.in", "r", stdin);
     freopen("A.out", "w", stdout);
     init();
     solve();
     return ;
 }

Problem A

Problem B GCD

题目大意

　　最小gcd生成树，点权1~n。

　　加速Kruskal的过程。边排序的时候把编号也作为关键字。

Code

 #include <iostream>
 #include <cstdlib>
 #include <cstdio>
 using namespace std;
 typedef bool boolean;

 int n;

 inline void init() {
     scanf("%d", &n);
 }

 int *uf;

 int find(int x) {
     return (uf[x] == x) ? (x) : (uf[x] = find(uf[x]));
 }

 long long res = ;
 inline void solve() {
     uf = new int[(n + )];
     for (int i = ; i <= n; i++)
         uf[i] = i;
     for (int i = n >> ; i; i--) {
         for (int j = i << ; j <= n; j += i) {
             if (find(i) != find(j)) {
                 res += i;
                 uf[find(i)] = find(j);
             }
         }
     }
     cout << res << endl;
 }

 int main() {
     freopen("B.in", "r", stdin);
     freopen("B.out", "w", stdout);
     init();
     solve();
     return ;
 }

Problem B

Problem C SEG

题目大意

　　平面的三角剖分。问边数的奇偶性。

　　mmp，最开始理解错题意了。是说咋觉得不可做。

　　可以求出凸包，设凸包上的点数和边数均为$m$。然后设顶点总数为$V$，边数为$E$，分割成的三角形的个数为$x$。

　　则根据欧拉公式有：$x + 1 + V - \frac{3x + m}{2} = 2$

　　解得：$x = 2n - m - 2$，由此得到边数。

　　另外这道题数据似乎对极角序不是很友好。

Code

 #include <algorithm>
 #include <iostream>
 #include <cstdlib>
 #include <cstdio>
 #include <cmath>
 #ifndef WIN32
 #define Auto "%lld"
 #else
 #define Auto "%I64d"
 #endif
 using namespace std;
 typedef bool boolean;

 #define ll long long

 typedef class Point {
     public:
         ll x, y;

         Point(ll x = 0.0, ll y = 0.0):x(x), y(y) {    }

         boolean operator < (Point b) const {
             if (x ^ b.x)
                 return x < b.x;
             return y < b.y;
         }
 }Point, Vector;

 Point operator - (Point a, Point b) {
     return Point(a.x - b.x, a.y - b.y);
 }

 ll cross(Point a, Point b) {
     return a.x * b.y - a.y * b.x;
 }

 int n, m;
 Point* ps;

 inline void init() {
     scanf("%d", &n);
     ps = new Point[(n + )];
     for (int i = ; i <= n; i++)
         scanf(Auto""Auto, &ps[i].x, &ps[i].y);
 }

 int st;
 Point* tp;

 inline void solve() {
     tp = new Point[(n + )];
     sort(ps + , ps + n + );
     for (int i = ; i <= n; i++) {
         while (st >  && cross(tp[st] - tp[st - ], ps[i] - tp[st]) <= )
             st--;
         tp[++st] = ps[i];
     }
     m += st - , st = ;
     for (int i = n; i; i--) {
         while (st >  && cross(tp[st] - tp[st - ], ps[i] - tp[st]) <= )
             st--;
         tp[++st] = ps[i];
     }
     m += st;

     int A = ((n << ) - m) - ;
     int E = ( * A + m) >> ;
     if (E & )
         puts("Alice");
     else
         puts("Bob");
 }

 int main() {
     init();
     solve();
     return ;
 }

Problem C