C.平分遊戲node
D.psd面試ios
感覺：dp仍是不會，雖然這是最簡單的dp。。。。。面試
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 const int maxn = 1500; 10 11 int n, dp[maxn][maxn]; 12 string s; 13 14 void LPS() { 15 memset(dp, 0, sizeof(dp)); 16 for (int i = 0; i < n; i++) dp[i][i] = 1; 17 for (int i = 1; i < n; i++) { 18 int tp = 0; 19 for (int j = 0; j + i < n; j++) { 20 if (s[j] == s[j + i]) tp = dp[j + 1][j + i - 1] + 2; 21 else tp = max(dp[j + 1][j + i], dp[j][j + i - 1]); 22 dp[j][j + i] = tp; 23 } 24 } 25 cout << n - dp[0][n - 1] << endl; 26 } 27 28 int main() 29 { 30 while (cin >> s) { 31 n = s.size(); 32 for (int i = 0; i < n; i++) if (s[i] >= 'A'&& s[i] <= 'Z') s[i] = tolower(s[i]); 33 LPS(); 34 } 35 return 0; 36 }
E.迴旋星空url
題解：枚舉中間的那個點，分別計算其它點到這個點的距離，再排序。那麼相同距離的就會排在一塊兒。spa
感覺：想到了枚舉中間點，但算錯了複雜度。。。。.net
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 typedef long long ll; 13 14 const int maxn = 1005; 15 16 int n, T; 17 18 struct node { 19 int x, y; 20 }p[maxn]; 21 22 /*struct mode { 23 int s, e, d; 24 bool operator<(const mode& i)const { 25 return d < i.d; 26 } 27 }q[maxn];*/ 28 29 inline int cad(node a, node b) { return (a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y); } 30 31 int main() 32 { 33 cin >> T; 34 while (T--) { 35 cin >> n; 36 for (int i = 1; i <= n; i++) cin >> p[i].x >> p[i].y; 37 38 ll ans = 0; 39 for (int i = 1; i <= n; i++) { 40 vector<int> d; 41 for (int j = 1; j <= n; j++) if (j != i)d.push_back(cad(p[i], p[j])); 42 sort(d.begin(), d.end()); 43 44 int cnt = 1; 45 for (int j = 1; j < d.size(); j++) { 46 if (d[j] == d[j - 1]) cnt++; 47 else { 48 ans += cnt * (cnt - 1); 49 cnt = 1; 50 } 51 } 52 ans += cnt * (cnt - 1); 53 } 54 if (ans == 0) cout << "WA" << endl; 55 else printf("%lld\n", ans); 56 57 } 58 return 0; 59 }
F.等式code
題解：假設x=n+a,y=n+b;代入方程得：n*n=a*b。因此若是n*n有p個因子，則ans=(p+1)/2;由於n*n的質因子和n的質因子是相同的，因此這裏分解n就好了。blog
感覺：網上的大佬就是6啊！不過這題是原題，poj2917。排序
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, T; 10 int num[64]; 11 12 int main() 13 { 14 cin >> T; 15 while (T--) { 16 cin >> n; 17 memset(num, 0, sizeof(num)); 18 int p = 0; 19 for (int i = 2; i*i <= n; i++) { 20 if (n % i) continue; 21 while (n % i == 0) { 22 num[p]++; 23 n /= i; 24 } 25 p++; 26 } 27 if (n != 1) { 28 num[p]++; 29 p++; 30 } 31 for (int i = 0; i < p; i++) num[i] *= 2; 32 int ans = 1; 33 for (int i = 0; i < p; i++) ans *= (num[i] + 1); 34 cout << (ans+1)/2 << endl; 35 } 36 return 0; 37 }
G.旋轉矩陣遊戲
題解：LR和RL等同沒有旋轉，因此旋轉到最後等價於只向左旋或只向右旋。
感覺：fuckkkkk！if-else結構居然寫掛了，比賽結束後真想找塊豆腐撞死。
比賽時寫的左旋：
1 /*左旋*/ 2 void print3() { 3 cout << m << " " << n << endl; 4 for (int i = m - 1; i >= 0; i--) { 5 for (int j = 0; j < n; j++) { 6 if (mp[j][i] == '|') mp[j][i] = '-'; 7 if (mp[j][i] == '-') mp[j][i] = '|'; //居然沒找出錯誤，orzzzzz! 8 cout << mp[j][i]; 9 } 10 cout << endl; 11 } 12 }
最後AC的代碼：
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 const int maxn = 2000; 10 11 int T, n, m; 12 char mp[100][100]; 13 14 string s; 15 16 void print1(){ 17 cout << n << " " << m << endl; 18 for (int i = 0; i < n; i++) { 19 for (int j = 0; j < m; j++) cout << mp[i][j]; 20 cout << endl; 21 } 22 } 23 /*右旋*/ 24 void print2() { 25 cout << m << " " << n << endl; 26 for (int i = 0; i < m; i++) { 27 for (int j = n - 1; j >= 0; j--) { 28 if (mp[j][i] == '|') cout << "-"; 29 else if (mp[j][i] == '-') cout << "|"; 30 else cout << mp[j][i]; 31 } 32 cout << endl; 33 } 34 } 35 /*左旋*/ 36 void print3() { 37 cout << m << " " << n << endl; 38 for (int i = m - 1; i >= 0; i--) { 39 for (int j = 0; j < n; j++) { 40 if (mp[j][i] == '|') cout << "-"; 41 else if (mp[j][i] == '-') cout << "|"; 42 else cout << mp[j][i]; 43 } 44 cout << endl; 45 } 46 } 47 /*左旋兩次*/ 48 void print4() { 49 cout << n << " " << m << endl; 50 for (int i = n - 1; i >= 0; i--) { 51 for (int j = m - 1; j >= 0; j--) cout << mp[i][j]; 52 cout << endl; 53 } 54 } 55 56 int main() 57 { 58 cin >> T; 59 while (T--) { 60 cin >> n >> m; 61 for (int i = 0; i < n; i++) 62 for (int j = 0; j < m; j++) cin >> mp[i][j]; 63 cin >> s; 64 65 int l = s.size(); 66 int p = 0, q = 0; 67 for (int i = 0; i < l; i++) { 68 if (s[i] == 'L') p++; 69 if (s[i] == 'R') q++; 70 } 71 72 if (p == q) print1(); 73 else if (p > q) { 74 p = (p - q) % 4; 75 if (p == 0) print1(); 76 else if (p == 1) print3(); 77 else if (p == 2) print4(); 78 else print2(); 79 } 80 else { 81 q = (q - p) % 4; 82 if (q == 0) print1(); 83 else if (q == 1) print2(); 84 else if (q == 2) print4(); 85 else print3(); 86 } 87 88 cout << endl; 89 } 90 return 0; 91 }
J.強迫症序列
題解：每次只能對n-1個數加一，等價於每次只能對1個數減一。並且每一個元素都相等的狀況只有一種。
1 #pragma warning(disable:4996) 2 #include<cstdio> 3 #include<cstring> 4 #include<iostream> 5 #include<algorithm> 6 using namespace std; 7 8 const int maxn = 1e5 + 5; 9 10 int n; 11 int a[maxn]; 12 13 int main() 14 { 15 int T; 16 while (cin >> T) { 17 while (T--) { 18 cin >> n; 19 for (int i = 1; i <= n; i++) scanf("%d", a + i); 20 sort(a + 1, a + n + 1); 21 int ans = 0; 22 for (int i = 1; i <= n; i++) ans += (a[i] - a[1]); 23 cout << ans << " " << ans + a[1] << endl; 24 } 25 } 26 return 0; 27 }
K.密碼
題解：找規律。
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 const int maxn = 100005; 10 11 int n, m; 12 char s[maxn]; 13 14 int main() 15 { 16 int T; 17 cin >> T; 18 while (T--) { 19 scanf("%d%s", &n, s); 20 m = strlen(s); 21 if (n == 1 || n >= m) { printf("%s\n", s); continue; } 22 23 for (int i = 0; i < n; i++) { 24 int t = i; 25 int p = 0; 26 while (t < m) { 27 cout << s[t]; 28 if (i == 0 || i == n - 1) { t += 2 * (n - 1); continue; } 29 if (p % 2) t += 2 * i; 30 else t += 2 * (n - i - 1); 31 p++; 32 } 33 } 34 cout << endl; 35 } 36 37 return 0; 38 }
L.用來做弊的藥水
題解：分類討論。
1 #pragma warning(disable:4996) 2 #include<cstdio> 3 #include<cstring> 4 #include<iostream> 5 #include<algorithm> 6 using namespace std; 7 8 const int maxn = 1e5 + 5; 9 10 int n; 11 int a[maxn]; 12 13 int main() 14 { 15 16 int T; 17 cin >> T; 18 while (T--) { 19 int x, a, y, b; 20 cin >> x >> a >> y >> b; 21 bool flag = false; 22 while (true) { 23 24 if (x == 1 && y == 1) { flag = true; break; } 25 if (x == 1 && y != 1) break; 26 if (x != 1 && y == 1) break; 27 28 if (a == b) { 29 if (x == y) { flag = true; break; } 30 else break; 31 } 32 else if (a < b) { 33 if (x < y || x % y) break; 34 else { 35 x = x / y; 36 b = b - a; 37 } 38 } 39 else { 40 if (y < x || y % x) break; 41 else { 42 y = y / x; 43 a = a - b; 44 } 45 } 46 if (x == y && a != b) break; 47 } 48 if (flag) cout << "Yes" << endl; 49 else cout << "No" << endl; 50 51 } 52 return 0; 53 }