2019 The 19th Zhejiang University Programming Contest

感想：

　　今天三个人的状态比昨天计院校赛的状态要好很多，然而三个人都慢热体质导致签到题wa了很多发。最后虽然跟大家题数一样(6题)，然而输在罚时。

　　只能说，水题还是刷得少，看到签到都没灵感实在不应该。

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题目链接：http://acm.zju.edu.cn/onlinejudge/showContestProblems.do?contestId=391

　　A：简单贪心，按高度sort一下就好了，这里用优先队列处理

 #include <cstdio>
 #include <queue>
 #include <algorithm>
 #include <functional>

 using namespace std;

 int M[], F[];

 int main(void)
 {
     int T;
     while (scanf("%d", &T) != EOF)
     {
         while (T--)
         {
             int n, m;
             priority_queue <int, vector<int>, greater <int>> fu, fd, mu, md;
             scanf("%d %d", &n, &m);
             for (int i = ; i < n; i++)
             {
                 scanf("%d", &M[i]);
             }
             for (int i = ; i < m; i++)
             {
                 scanf("%d", &F[i]);
             }
             for (int i = ; i < n; i++)
             {
                 int flag;
                 scanf("%d", &flag);
                 if (flag)
                 {
                     mu.push(M[i]);
                 }
                 else
                 {
                     md.push(M[i]);
                 }
             }
             for (int i = ; i < m; i++)
             {
                 int flag;
                 scanf("%d", &flag);
                 if (flag)
                 {
                     fu.push(F[i]);
                 }
                 else
                 {
                     fd.push(F[i]);
                 }
             }
             int ans = ;
             // fu <-> md
             while (!fu.empty() && !md.empty())
             {
                 if (fu.top() < md.top())
                 {
 //                  printf("pop: %d %d\n", fu.top(), md.top());
                     fu.pop();
                     md.pop();
                     ans++;
                 }
                 else
                 {
 //                  printf("pop: %d\n", md.top());
                     md.pop();
                 }
             }
             // fd <-> mu
             while (!mu.empty() && !fd.empty())
             {
                 if (mu.top() < fd.top())
                 {
 //                  printf("pop: %d %d\n", mu.top(), fd.top());
                     mu.pop();
                     fd.pop();
                     ans++;
                 }
                 else
                 {
 //                  printf("pop: %d\n", fd.top());
                     fd.pop();
                 }
             }
             printf("%d\n", ans);
         }
     }
     return ;
 }

　　B：找规律，显然是不停把n除二加起来，高精就用java

 import java.util.*;
 import java.lang.*;
 import java.math.BigInteger;

 public class Main {

     public static void main(String[] args) {
         Scanner cin=new Scanner(System.in);
         while (cin.hasNextInt())
         {
             int t=cin.nextInt();
             while (t-->0){
                 BigInteger ans=BigInteger.ZERO;
                 BigInteger x=cin.nextBigInteger();
                 while (!x.equals(BigInteger.ONE)){
                     x=x.divide(BigInteger.valueOf(2));
                     ans=ans.add(x);
                 }
                 System.out.println(ans);
             }
         }
     }
 }

　　C：模拟

 #include <cstdio>

 char t[][];
 bool vis[][];

 const int p3[] = {, , , , , };

 const int movement[][] =
 {
     {, }, // DOWN
     {, },  // RIGHT
     {-, }, // UP
     {, -} // LEFT
 };

 void init(int n, int m)
 {
     for (int i = ; i < n; i++)
     {
         for (int j = ; j < m; j++)
         {
             vis[i][j] = false;
         }
     }
 }

 int solve(void)
 {
     int ans = ;
     int n, m;
     int a, b;
     long long int k;
     scanf("%d %d", &n, &m);
     scanf("%d %d %lld", &a, &b, &k);
     a--;
     b--;
     char cmd[];
     scanf(" %s", cmd);
     init(n, m);
     for (int i = ; i < n; i++)
     {
         scanf(" %s", t[i]);
         for (int j = ; j < m; j++)
         {
             t[i][j] -= '';
         }
     }
     while (k--)
     {
         int x = p3[] * t[a][b]
                 + p3[] * t[a - ][b]
                 + p3[] * t[a + ][b]
                 + p3[] * t[a][b - ]
                 + p3[] * t[a][b + ];
         if (vis[a][b])
         {
             return ans;
         }
         vis[a][b] = true;
         if (cmd[x] == 'D')
         {
             int na = a + movement[][], nb = b + movement[][];
             if (t[na][nb] == ) return ans;
             else a = na, b = nb;
         }
         else if (cmd[x] == 'R')
         {
             int na = a + movement[][], nb = b + movement[][];
             if (t[na][nb] == ) return ans;
             else a = na, b = nb;
         }
         else if (cmd[x] == 'U')
         {
             int na = a + movement[][], nb = b + movement[][];
             if (t[na][nb] == ) return ans;
             else a = na, b = nb;
         }
         else if (cmd[x] == 'L')
         {
             int na = a + movement[][], nb = b + movement[][];
             if (t[na][nb] == ) return ans;
             else a = na, b = nb;
         }
         else if (cmd[x] == 'P')
         {
             if (t[a][b] == )
             {
                 t[a][b] = ;
                 ans++;
                 init(n, m);
             }
         }
         else if (cmd[x] == 'I')
         {
             return ans;
         }
     }
     return ans;
 }

 int main(void)
 {
     int T;
     while (scanf("%d", &T) != EOF)
     {
         while (T--)
         {
             printf("%d\n", solve());
         }
     }
     return ;
 }

　　D：上一题的人工智能版，要你构造特定程序捡垃圾。方法是走回字形，比如我们选定顺时针方向走，那么当我们走到左边靠墙位置的时候，如果右手边有垃圾，那么我们往右走一格再继续往上走。如果走到地图中间位置(四周没垃圾)就往上走。如果机器人走了连续相同方向n次就让机器人“抖动”一下(属实人工智能调参)。然而cy他们队就是A了(神仙啊

　　E：从右往左扫一次就好了，队友没开LL导致wa一发要批评

 #include <cstdio>

 long long int a[], b[];

 int main(void)
 {
     int T;
     while (scanf("%d", &T) != EOF)
     {
         while (T--)
         {
             int n;
             scanf("%d", &n);
             for (int i = ; i < n; i++)
             {
                 scanf("%lld", &a[i]);
             }
             for (int i = ; i < n; i++)
             {
                 scanf("%lld", &b[i]);
             }
             bool flag = true;
             for (int i = n - ; i >= ; i--)
             {
                 if (b[i] >= a[i])
                 {
                     if (i)
                     {
                         b[i - ] += b[i] - a[i];
                     }
                 }
                 else
                 {
                     flag = false;
                     break;
                 }
             }
             printf(flag ? "Yes\n" : "No\n");
         }
     }
     return ;
 }

　　F：神仙题

　　G：非常简单的贪心

 #include <cstdio>
 #include <queue>
 #include <algorithm>
 #include <functional>

 using namespace std;

 int M[], F[];

 int main(void)
 {
     int T;
     while (scanf("%d", &T) != EOF)
     {
         while (T--)
         {
             int n, k;
             scanf("%d %d", &n, &k);
             priority_queue <long long int> pos, neg;
             for (int i = ; i < n; i++)
             {
                 int tmp;
                 scanf("%d", &tmp);
                 if (tmp > )
                 {
                     pos.push(tmp);
                 }
                 else
                 {
                     neg.push(-tmp);
                 }
             }
             long long int ans = , maxn = ;
             while (!pos.empty())
             {
                 ans +=  * pos.top();
                 maxn = max(pos.top(), maxn);
                 int cnt = k - ;
                 pos.pop();
                 while (cnt-- && !pos.empty())
                 {
                     pos.pop();
                 }
             }
             while (!neg.empty())
             {
                 ans +=  * neg.top();
                 maxn = max(neg.top(), maxn);
                 int cnt = k - ;
                 neg.pop();
                 while (cnt-- && !neg.empty())
                 {
                     neg.pop();
                 }
             }
             printf("%lld\n", ans - maxn);
         }
     }
     return ;
 }

　　H：救公主，边双相关的题目(然而队友最后没撸出来)

　　 I：要求逆元的看不懂的题目

　　J：签到(wa了7次，三个人都没睡醒。这里我马上想到了可以输出n×2和n×3，然而忘记特判n==1的情况……)

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>

 using namespace std;

 typedef long long ll;

 ll read()
 {
     ll x = ; char c = getchar(); ll flag = ;
     while (c < '' || c > '')
     {
         if (c == '-')
         {
             flag = -;
         }
         c = getchar();
     }
     while (c >= '' && c <= '')x = x * 10ll + c - '', c = getchar();
     return x;
 }

 int main()
 {
     ll T;
     while (scanf("%lld", &T) == )
     {
         while (T--)
         {
             ll x;
             x = read();
             if (x %  == )
             {
                 printf("4 %lld\n",  + x);
             }
             else
             {
                 printf("15 %lld\n",  + x);
             }
         }
     }

     return ;
 }