A. Tritonic Iridescence

題解：分類討論。註意題目要求，至少有兩種方案。

 1 #pragma warning(disable:4996)
 2 #include<cstdio>
 3 #include<string>
 4 #include<cstring>
 5 #include<iostream>
 6 #include<algorithm>
 7 using namespace std;
 8 
 9 int n, m;
10 string s;
11 
12 int main()
13 {
14     while
 
 (cin >> n) {
15         cin >> s;
16         int cnt = 0;
17         for (int i = 0; i < n; i++) if (s[i] == ‘?‘) cnt++;
18         if (!cnt) printf("No\n");
19         else {
20             bool flag1 = false;
21             bool flag2 = true;
22             for (int i = 1; i < n; i++) {
 
23                 if (s[i] == s[i - 1] && s[i] == ‘?‘) flag1 = true;
24                 if (s[i] == s[i - 1] && s[i] != ‘?‘) flag2 = false;
25             }
26             if (!flag2) printf("No\n");
27             else {
28                 if (flag1) printf("Yes\n");
29                 else
 
 {
30                     bool flag3 = false;
31                     for (int i = 1; i < n - 1; i++) {
32                         if (s[i] == ‘?‘) {
33                             if (s[i - 1] == s[i + 1]) flag3 = true;
34                         }
35                     }
36                     if (s[0] == ‘?‘ || s[n - 1] == ‘?‘) flag3 = true;
37                     if (!flag3) printf("No\n");
38                     else printf("Yes\n");
39                 }
40             }
41         }
42     }
43     return 0;
44 }

B. Mystical Mosaic

題解：對於同一列的"#"所在行應該是對稱的。

 1 #pragma warning(disable:4996)
 2 #include<map>
 3 #include<vector>
 4 #include<stack>
 5 #include<queue>
 6 #include<cstdio>
 7 #include<string>
 8 #include<cstring>
 9 #include<iostream>
10 #include<algorithm>
11 using namespace std;
12 
13 const int maxn = 60;
14 
15 int n, m;
16 int mp[maxn][maxn];
17 
18 int main()
19 {
20     scanf("%d%d", &n, &m);
21     getchar();
22     for (int i = 1; i <= n; i++) {
23         for (int j = 1; j <= m; j++)
24             mp[i][j] = (getchar() == ‘#‘);
25         getchar();
26     }
27     for (int i = 1; i <= n; i++) {
28         for (int j = i + 1; j <= n; j++) {
29             bool flag1 = true, flag2 = true;
30             for (int k = 1; k <= m; k++) {
31                 if (mp[i][k] != mp[j][k]) flag1 = false;
32                 if (mp[i][k] && mp[j][k]) flag2 = false;
33             }
34             if (!flag1 && !flag2) { puts("No"); return 0; }
35         }
36     }
37     puts("Yes"); return 0;
38 }

C. Three-level Laser

題解：分析式子，當 E_j 選定後，i=j-1，所以只需要考慮 E_k ,顯然 k 的位置離 j 越遠答案越優。

註意：精度。

 1 #pragma warning(disable:4996)
 2 #include<map>
 3 #include<queue>
 4 #include<string>
 5 #include<vector>
 6 #include<cstdio>
 7 #include<cstring>
 8 #include<iostream>
 9 #include<algorithm>
10 using namespace std;
11 typedef long long ll;
12 
13 const int maxn = 1e5 + 5;
14 
15 
16 int n, U;
17 int a[maxn];
18 
19 int main()
20 {
21     while (scanf("%d%d", &n, &U) != EOF) {
22         for (int i = 1; i <= n; i++) scanf("%d", a + i);
23         double ans = 0;
24         for (int i = 2; i < n; i++) {
25             int p = lower_bound(a + 1, a + n + 1, a[i - 1] + U) - (a);
26             
27             if (a[p] != a[i - 1] + U) p--;
28             if (p == i) continue;
29 
30             double tp = 1.0*(a[p] - a[i]) / (a[p] - a[i - 1]);
31             ans = max(ans, tp);
32         }
33         if (!ans) printf("-1\n");
34         else printf("%.12lf\n", ans);
35     }
36     return 0;
37 }

Codeforces Round #472 (rated, Div. 2, based on VK Cup 2018 Round 2)