摘要：A1 = ?Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7535 Accepted Submission(s): 4675 Problem Description 有如下方程：Ai = (Ai-1 + Ai+1)/2 - Ci (... 阅读全文

摘要：核反应堆Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15706 Accepted Submission(s): 7036 Problem Description 某核反应堆有两类事件发生：高能质点碰击核子时，质点被吸收，放出3个高... 阅读全文

摘要：数塔Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 39043 Accepted Submission(s): 23246 Problem Description 在讲述DP算法的时候，一个经典的例子就是数塔问题，它是这样描述的：有如... 阅读全文

摘要：简易版之最短距离Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 17644 Accepted Submission(s): 7904 Problem Description 寒假的时候，ACBOY要去拜访很多朋友，恰巧他所有朋友的家都... 阅读全文

摘要：手机短号Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 28591 Accepted Submission(s): 17478 Problem Description 大家都知道，手机号是一个11位长的数字串，同时，作为学生，还可以申... 阅读全文

摘要：夹角有多大IITime Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10656 Accepted Submission(s): 5510 Problem Description 这次xhd面临的问题是这样的：在一个平面内有两个点，求两个点分... 阅读全文

摘要：A|B? Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 18605 Accepted Submission(s): 14122 Problem 阅读全文

摘要：Max Num Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 22270 Accepted Submission(s): 13029 Problem Description There are some students in a ... 阅读全文

摘要：Fibbonacci Number Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 22530 Accepted Submission(s): 10375 Problem Description Your objective for ... 阅读全文

摘要：Coin Change Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 18271 Accepted Submission(s): 6291 Pr 阅读全文

摘要：根据公式计算卡特兰数，始终出现越界，自己无法解决 根据递推公式 动态规划 阅读全文

摘要：#include #include using namespace std; int main() { int n,k; double gpa,s,credit,score; string name; bool flag = true; cin >> n; while (n--) { cin >> k; ... 阅读全文

摘要：超时 a1 = k an = k + n - 1 M = n*(2k+n-1)/2 解的 k = M/n - (n-1)/2，这个不能这样写，要写在一起 (2M-n*n-n)/(2*n)。否则对于30/4 - 3/2，会忽略这样的结果，如果先通分当然也就可以。 超时 利用a1>=1这个条件可以减少循 阅读全文

摘要：16进制的加法 阅读全文

摘要：#include #include #include using namespace std; struct node { double a, b, c, d; node(){} void add(double a, double b, double c, double d){ this->a = a; this->b = b; this->c = c; thi... 阅读全文

摘要：#include #include using namespace std; bool islower(char ch) { if (ch >= 'a' && ch = 'A' && ch > T; while (T--) { cin >> ch >> y; sum = y; if (islower... 阅读全文

摘要：#include #include #include using namespace std; /* 无穷多次操作，第n灯的状态 */ int main() { long long n; while (cin >> n) { int count = 0; for (int i = 1; i <= n; i++) ... 阅读全文

摘要：#include #include using namespace std; int main() { int n, m; while (cin >> m >> n) { for (int i = 0; i <= n+1; i++) { if (i == 0 || i == n+1) ... 阅读全文

摘要：#include #include #include #include using namespace std; void toBit(int n) { vector vec; while (n > 0) { if (n % 2 == 0) { vec.push_back(0); } ... 阅读全文

摘要：太难，百度一下 阅读全文