题解：分类讨论。注意题目要求，至少有两种方案。
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 "); 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 "); 27 else { 28 if (flag1) printf("Yes "); 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 "); 38 else printf("Yes "); 39 } 40 } 41 } 42 } 43 return 0; 44 }
题解：对于同一列的"#"所在行应该是对称的。
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 }
题解：分析式子，当 Ej 选定后，i=j-1，所以只需要考虑 Ek ,显然 k 的位置离 j 越远答案越优。
注意：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 "); 34 else printf("%.12lf ", ans); 35 } 36 return 0; 37 }