题目传送门:http://acm.hdu.edu.cn/showproblem.php?pid=6438 Buy and Resell Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1233    Accepted Submission(s): 407 Problem Description The Power Cube is used…

Buy and Resell Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1126    Accepted Submission(s): 359 Problem Description The Power Cube is used as a stash of Exotic Power. There are n cities numbe…

题目传送门:http://acm.hdu.edu.cn/showproblem.php?pid=6447 YJJ's Salesman Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 919    Accepted Submission(s): 290 Problem Description YJJ is a salesman who h…

Tree and Permutation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 619    Accepted Submission(s): 214 Problem Description There are N vertices connected by N−1 edges, each edge has its own len…

Problem Descripton Two planets named Haha and Xixi in the universe and they were created with the universe beginning. There is 73 days in Xixi a year and 137 days in Haha a year. Now you know the days N after Big Bang, you need to answer whether it i…

给定的\(p\)是素数,要求给定一个加法运算表和乘法运算表,使\((m+n)^p = m^p +n^p(0 \leq m,n < p)\). 因为给定的p是素数,根据费马小定理得 \((m+n)^{p-1} \equiv 1(mod\ p)\) 因此,\((m+n)^{p} \equiv m + n\ (mod\ p)\), 同时,\(m^p + n^p \equiv m + n\ (mod\ p)\). 所以在模p意义下,\((m+n)^p = m^p +n^p(0\leq m,n < p)…

http://acm.hdu.edu.cn/showproblem.php?pid=6440 题意:让你重新定义任意一对数的乘法和加法结果(输出乘法口诀表和加法口诀表),使得m^p+n^p==(m+n)^p(p为质数),并且存在一个0<q<p使得 q^k(0<k<p)取遍1~p-1的所有值,并且该运算是封闭的(exists an integer q(0<q<p) to make the set {qk|0<k<p,k∈Z} equal to {k|0<…

Neko's loop Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1386 Accepted Submission(s): 316 Problem Description Neko has a loop of size n. The loop has a happy value ai on the i−th(0≤i≤n−1) grid.…

A - Buy and Resell 题意:给出n个交易点,每次能够选择买或者卖,求获得最大利润 思路:维护两个优先队列,一个是卖,一个是替换,当价格差相同时,优先替换,因为次数要最少 #include <bits/stdc++.h> using namespace std; #define ll long long #define N 100010 int t, n; ll arr[N]; priority_queue <ll, vector <ll>, greater &…

Find Integer Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 6597    Accepted Submission(s): 1852Special Judge Problem Description people in USSS love math very much, and there is a famous math…