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 }
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 }
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 }
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 }
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 }
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 }
