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。

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

 1 #include <cstdio>
 2 #include <queue>
 3 #include <algorithm>
 4 #include <functional>
 5 
 6 using namespace std;
 7 
 8 int M[100005], F[100006];
 9 
10 int main(void)
11 {
12     int T;
13     while (scanf("%d", &T) != EOF)
14     {
15         while (T--)
16         {
17             int n, m;
18             priority_queue <int, vector<int>, greater <int>> fu, fd, mu, md;
19             scanf("%d %d", &n, &m);
20             for (int i = 0; i < n; i++)
21             {
22                 scanf("%d", &M[i]);
23             }
24             for (int i = 0; i < m; i++)
25             {
26                 scanf("%d", &F[i]);
27             }
28             for (int i = 0; i < n; i++)
29             {
30                 int flag;
31                 scanf("%d", &flag);
32                 if (flag)
33                 {
34                     mu.push(M[i]);
35                 }
36                 else
37                 {
38                     md.push(M[i]);
39                 }
40             }
41             for (int i = 0; i < m; i++)
42             {
43                 int flag;
44                 scanf("%d", &flag);
45                 if (flag)
46                 {
47                     fu.push(F[i]);
48                 }
49                 else
50                 {
51                     fd.push(F[i]);
52                 }
53             }
54             int ans = 0;
55             // fu <-> md
56             while (!fu.empty() && !md.empty())
57             {
58                 if (fu.top() < md.top())
59                 {
60 //                  printf("pop: %d %d\n", fu.top(), md.top());
61                     fu.pop();
62                     md.pop();
63                     ans++;
64                 }
65                 else
66                 {
67 //                  printf("pop: %d\n", md.top());
68                     md.pop();
69                 }
70             }
71             // fd <-> mu
72             while (!mu.empty() && !fd.empty())
73             {
74                 if (mu.top() < fd.top())
75                 {
76 //                  printf("pop: %d %d\n", mu.top(), fd.top());
77                     mu.pop();
78                     fd.pop();
79                     ans++;
80                 }
81                 else
82                 {
83 //                  printf("pop: %d\n", fd.top());
84                     fd.pop();
85                 }
86             }
87             printf("%d\n", ans);
88         }
89     }
90     return 0;
91 }

View Code

　　B：找规律，显然是不停把n除二加起来，高精就用java

 1 import java.util.*;
 2 import java.lang.*;
 3 import java.math.BigInteger;
 4 
 5 public class Main {
 6 
 7     public static void main(String[] args) {
 8         Scanner cin=new Scanner(System.in);
 9         while (cin.hasNextInt())
10         {
11             int t=cin.nextInt();
12             while (t-->0){
13                 BigInteger ans=BigInteger.ZERO;
14                 BigInteger x=cin.nextBigInteger();
15                 while (!x.equals(BigInteger.ONE)){
16                     x=x.divide(BigInteger.valueOf(2));
17                     ans=ans.add(x);
18                 }
19                 System.out.println(ans);
20             }
21         }
22     }
23 }

View Code

　　C：模拟

  1 #include <cstdio>
  2 
  3 char t[2002][2002];
  4 bool vis[2002][2002];
  5 
  6 const int p3[] = {1, 3, 9, 27, 81, 243};
  7 
  8 const int movement[4][2] =
  9 {
 10     {1, 0}, // DOWN
 11     {0, 1},  // RIGHT
 12     {-1, 0}, // UP
 13     {0, -1} // LEFT
 14 };
 15 
 16 void init(int n, int m)
 17 {
 18     for (int i = 0; i < n; i++)
 19     {
 20         for (int j = 0; j < m; j++)
 21         {
 22             vis[i][j] = false;
 23         }
 24     }
 25 }
 26 
 27 int solve(void)
 28 {
 29     int ans = 0;
 30     int n, m;
 31     int a, b;
 32     long long int k;
 33     scanf("%d %d", &n, &m);
 34     scanf("%d %d %lld", &a, &b, &k);
 35     a--;
 36     b--;
 37     char cmd[300];
 38     scanf(" %s", cmd);
 39     init(n, m);
 40     for (int i = 0; i < n; i++)
 41     {
 42         scanf(" %s", t[i]);
 43         for (int j = 0; j < m; j++)
 44         {
 45             t[i][j] -= '0';
 46         }
 47     }
 48     while (k--)
 49     {
 50         int x = p3[4] * t[a][b]
 51                 + p3[3] * t[a - 1][b]
 52                 + p3[2] * t[a + 1][b]
 53                 + p3[1] * t[a][b - 1]
 54                 + p3[0] * t[a][b + 1];
 55         if (vis[a][b])
 56         {
 57             return ans;
 58         }
 59         vis[a][b] = true;
 60         if (cmd[x] == 'D')
 61         {
 62             int na = a + movement[0][0], nb = b + movement[0][1];
 63             if (t[na][nb] == 1) return ans;
 64             else a = na, b = nb;
 65         }
 66         else if (cmd[x] == 'R')
 67         {
 68             int na = a + movement[1][0], nb = b + movement[1][1];
 69             if (t[na][nb] == 1) return ans;
 70             else a = na, b = nb;
 71         }
 72         else if (cmd[x] == 'U')
 73         {
 74             int na = a + movement[2][0], nb = b + movement[2][1];
 75             if (t[na][nb] == 1) return ans;
 76             else a = na, b = nb;
 77         }
 78         else if (cmd[x] == 'L')
 79         {
 80             int na = a + movement[3][0], nb = b + movement[3][1];
 81             if (t[na][nb] == 1) return ans;
 82             else a = na, b = nb;
 83         }
 84         else if (cmd[x] == 'P')
 85         {
 86             if (t[a][b] == 2)
 87             {
 88                 t[a][b] = 0;
 89                 ans++;
 90                 init(n, m);
 91             }
 92         }
 93         else if (cmd[x] == 'I')
 94         {
 95             return ans;
 96         }
 97     }
 98     return ans;
 99 }
100 
101 int main(void)
102 {
103     int T;
104     while (scanf("%d", &T) != EOF)
105     {
106         while (T--)
107         {
108             printf("%d\n", solve());
109         }
110     }
111     return 0;
112 }

View Code

　　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

 1 #include <cstdio>
 2 
 3 long long int a[200], b[200];
 4 
 5 int main(void)
 6 {
 7     int T;
 8     while (scanf("%d", &T) != EOF)
 9     {
10         while (T--)
11         {
12             int n;
13             scanf("%d", &n);
14             for (int i = 0; i < n; i++)
15             {
16                 scanf("%lld", &a[i]);
17             }
18             for (int i = 0; i < n; i++)
19             {
20                 scanf("%lld", &b[i]);
21             }
22             bool flag = true;
23             for (int i = n - 1; i >= 0; i--)
24             {
25                 if (b[i] >= a[i])
26                 {
27                     if (i)
28                     {
29                         b[i - 1] += b[i] - a[i];
30                     }
31                 }
32                 else
33                 {
34                     flag = false;
35                     break;
36                 }
37             }
38             printf(flag ? "Yes\n" : "No\n");
39         }
40     }
41     return 0;
42 }

View Code

　　F：神仙题

　　G：非常简单的贪心

 1 #include <cstdio>
 2 #include <queue>
 3 #include <algorithm>
 4 #include <functional>
 5 
 6 using namespace std;
 7 
 8 int M[100005], F[100006];
 9 
10 int main(void)
11 {
12     int T;
13     while (scanf("%d", &T) != EOF)
14     {
15         while (T--)
16         {
17             int n, k;
18             scanf("%d %d", &n, &k);
19             priority_queue <long long int> pos, neg;
20             for (int i = 0; i < n; i++)
21             {
22                 int tmp;
23                 scanf("%d", &tmp);
24                 if (tmp > 0)
25                 {
26                     pos.push(tmp);
27                 }
28                 else
29                 {
30                     neg.push(-tmp);
31                 }
32             }
33             long long int ans = 0, maxn = 0;
34             while (!pos.empty())
35             {
36                 ans += 2 * pos.top();
37                 maxn = max(pos.top(), maxn);
38                 int cnt = k - 1;
39                 pos.pop();
40                 while (cnt-- && !pos.empty())
41                 {
42                     pos.pop();
43                 }
44             }
45             while (!neg.empty())
46             {
47                 ans += 2 * neg.top();
48                 maxn = max(neg.top(), maxn);
49                 int cnt = k - 1;
50                 neg.pop();
51                 while (cnt-- && !neg.empty())
52                 {
53                     neg.pop();
54                 }
55             }
56             printf("%lld\n", ans - maxn);
57         }
58     }
59     return 0;
60 }

View Code

　　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……)

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 
 5 using namespace std;
 6 
 7 typedef long long ll;
 8 
 9 ll read()
10 {
11     ll x = 0; char c = getchar(); ll flag = 1;
12     while (c < '0' || c > '9')
13     {
14         if (c == '-')
15         {
16             flag = -1;
17         }
18         c = getchar();
19     }
20     while (c >= '0' && c <= '9')x = x * 10ll + c - '0', c = getchar();
21     return x;
22 }
23 
24 int main()
25 {
26     ll T;
27     while (scanf("%lld", &T) == 1)
28     {
29         while (T--)
30         {
31             ll x;
32             x = read();
33             if (x % 2 == 0)
34             {
35                 printf("4 %lld\n", 4 + x);
36             }
37             else
38             {
39                 printf("15 %lld\n", 15 + x);
40             }
41         }
42     }
43 
44     return 0;
45 }