Brute Force Sorting Time Limit: 1 Sec  Memory Limit: 128 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=6215 Description Beerus needs to sort an array of N integers. Algorithms are not Beerus's strength. Destruction is what he excels. He can destr…

Apple Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 982    Accepted Submission(s): 323 Problem Description Apple is Taotao's favourite fruit. In his backyard, there are three apple trees with…

The Dominator of Strings Time Limit: 3000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description Here you have a set of strings. A dominator is a string of the set dom…

HDU 6206 Apple 题意: 给出四个点的坐标(每个点的坐标值小于等于1,000,000,000,000),问最后一个点是否在前三个点组成的三角形的外接圆内,是输出Accept,否输出Rejected 解题思路: 题意很好理解,就是判断一个点是否在一个圆内,或者说一个点到圆心的距离是否大于半径,关键是大整数和精度问题,题解中给出了java的解法. 设x0,y0为圆心,有圆心到三个顶点的距离相等,列出如下两个式子: (x1-x0)*(x1-x0)+(y1-y0)*(y1-y0) = (x2…

题目链接  Qingdao Problem C AC自动机还不会,暂时暴力水过. #include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i(a); i <= (b); ++i) #define dec(i, a, b) for (int i(a); i >= (b); --i) const int N = 1e5 + 10; string s[N]; int T; int n; int…

hdu 6197 题意:给定一个数组,问删掉k个字符后数组是否能不减或者不增,满足要求则是magic array,否则不是. 题解:队友想的思路,感觉非常棒!既然删掉k个后不增或者不减,那么就先求数组的最长不下降子序列的长度l1和最长不上升子序列的长度l2,若l1>=n-k||l2>=n-k,则满足要求. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm&g…

思路:找规律发现这个数是斐波那契第2*k+3项-1,数据较大矩阵快速幂搞定.   快速幂入门第一题QAQ #include <stdio.h> #include <stdlib.h> #include <cmath> #include <string.h> #include <iostream> #include <algorithm> #include <queue> #include <vector> #i…

A : Apple 题意:给出三个点,以及另一个点,求最后一个点是否在三个点的外接圆里面,如果在或者在边界上,输出“Rejected”,否则输出"Accepted" 思路:先求一个外接圆,得到圆心和半径,再判一下,注意精度问题,用JAVA import java.math.BigDecimal; import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner i…

2017-09-17 17:12:11 writer:pprp 找规律,质数只有是两个相邻的立方数的差才能形成,公式就是3 * n * (n + 1) +1, 判断读入的数是不是满足 这次依然只是做了两道签到题,其他题做了,但是因为没有想到好的算法,所以没有通过,唯一值得高兴的是这两个都是我自己做的 #include <iostream> #include <cstdio> #include <cmath> using namespace std; typedef lo…

2017-09-17 13:28:04 writer:pprp 签到题:1008 Chinese Zodiac #include <iostream> #include <string> #include <cstdio> #include <algorithm> #include <cmath> using namespace std; ] = {"rat", "ox", "tiger"…