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：用优先队列处理一下就好了
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：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：纯模拟
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)
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 }
View Code