B. Clique Problem time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output The clique problem is one of the most well-known NP-complete problems. Under some simplification it can be formulated as f…

传送门 D. Clique Problem time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output The clique problem is one of the most well-known NP-complete problems. Under some simplification it can be formulated…

B - Clique Problem 题目大意:给你坐标轴上n个点,每个点的权值为wi,两个点之间有边当且仅当 |xi - xj| >= wi + wj, 问你两两之间都有边的最大点集的大小. 思路:其实问题能转换为一堆线段问你最多能挑出多少个线段使其两两不相交.. 对于一个点来说可以变成[x - wi, x + wi]这样一条线段.. 然后贪心取就好啦. #include<bits/stdc++.h> #define LL long long #define fi first #def…

B. Clique Problem time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output The clique problem is one of the most well-known NP-complete problems. Under some simplification it can be formulated as f…

Codeforces Round #296 (Div. 1)C. Data Center Drama Time Limit: 2 Sec  Memory Limit: 256 MBSubmit: xxx  Solved: 2xx 题目连接 http://codeforces.com/contest/528/problem/C Description The project of a data center of a Big Software Company consists of n compu…

题目链接:Codeforces Round #367 (Div. 2) C. Hard problem 题意: 给你一些字符串,字符串可以倒置,如果要倒置,就会消耗vi的能量,问你花最少的能量将这些字符串排成字典序 题解: 当时1点过头太晕了,看错题了,然后感觉全世界都会,就我不会,- -!结果就是一个简单的DP, 设dp[i][0]表示第i个字符串不反转的情况,dp[i][1]表示第i个字符串反转的情况 状态转移方程看代码 #include<bits/stdc++.h> #define F(…

Codeforces Round #603 (Div. 2) A. Sweet Problem A. Sweet Problem time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You have three piles of candies: red, green and blue candies: the first pile…

Hard problem 题目链接: http://codeforces.com/contest/706/problem/C Description Vasiliy is fond of solving different tasks. Today he found one he wasn't able to solve himself, so he asks you to help. Vasiliy is given n strings consisting of lowercase Engl…

http://codeforces.com/contest/528/problem/E 先来吐槽一下,一直没机会进div 1, 马力不如当年, 这场题目都不是非常难,div 2 四道题都是水题! 题目大意:给n条直线,保证直线两两不平行,保证三条直线不公点.然后,随机挑三条直线,构成一个三角形,问挑出的三角形的面积的期望. 换句话说,就是算出全部三角形的面积和.再除以三角形的数量.后者是C(n,3), 关键是三角形面积和怎样计算. 下面给我我的思路:O(n^2) 基于三角形的向量表示法, S(A…

题目连接:http://codeforces.com/contest/688/problem/C 题意:给你一些边,问你能否构成一个二分图 题解:二分图:二分图又称作二部图,是图论中的一种特殊模型. 设G=(V,E)是一个无向图,如果顶点V可分割为两个互不相交的子集(A,B),并且图中的每条边(i,j)所关联的两个顶点i和j分别属于这两个不同的顶点集(i in A,j in B),则称图G为一个二分图. 直接上一个DFS就搞定了 #include<cstdio> #include<set…

C. NP-Hard Problem 题目连接: http://www.codeforces.com/contest/688/problem/C Description Recently, Pari and Arya did some research about NP-Hard problems and they found the minimum vertex cover problem very interesting. Suppose the graph G is given. Subs…

链接: https://codeforces.com/contest/1263/problem/A 题意: You have three piles of candies: red, green and blue candies: the first pile contains only red candies and there are r candies in it, the second pile contains only green candies and there are g ca…

C. Tourist Problem time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Iahub is a big fan of tourists. He wants to become a tourist himself, so he planned a trip. There are n destinations on a…

C. Tourist Problem time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Iahub is a big fan of tourists. He wants to become a tourist himself, so he planned a trip. There are n destinations on a…

C. NP-Hard Problem time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Recently, Pari and Arya did some research about NP-Hard problems and they found the minimum vertex cover problem very int…

A. Sweet Problem the first pile contains only red candies and there are r candies in it, the second pile contains only green candies and there are g candies in it, the third pile contains only blue candies and there are b candies in it. Each day Tany…

解题思路:就是求数 n 对应的二进制数中有多少个 1 #include <iostream> #include<cstdio> using namespace std; int main(){ int n; cin>>n; ; // while(n){//这也是一种好的方法 // n = n&(n-1); // ++ans; // } while(n){ ) ++ans; n>>=; } cout<<ans<<endl; ;…

screen 尺寸为a:b video 尺寸为 c:d 如果a == c 则 面积比为 cd/ab=ad/cb (ad < cb) 如果b == d 则 面积比为 cd/ab=cb/ad  (cb < ad) 如果不相等时 如果a/b > c/d,则ad/bd > cb/db 则(ad > cb) screen尺寸可为 ad:bd, video的尺寸可为 cb:db 面积比为:cb*db/ad*bd = cb/ad (ad > cb) 如果a/b < c/d,则a…

题目传送门 /* 无算法 三种可能:1.交换一对后正好都相同,此时-2 2.上面的情况不可能,交换一对后只有一个相同,此时-1 3.以上都不符合,则不交换,-1 -1 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstring> #include <string> #include <map>…

题目传送门 /* 水题 a或b成倍的减 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstring> #include <string> #include <map> #include <set> #include <vector> #include <set&…

A. Playing with Paper 如果a是b的整数倍,那么将得到a/b个正方形,否则的话还会另外得到一个(b, a%b)的长方形. 时间复杂度和欧几里得算法一样. #include <iostream> #include <cstdio> using namespace std; //const int maxn = ; int main() { //freopen("in.txt", "r", stdin); ; scanf(&q…

A. Playing with Paper One day Vasya was sitting on a not so interesting Maths lesson and making an origami from a rectangular a mm  ×  b mm sheet of paper (a > b). Usually the first step in making an origami is making a square piece of paper from the…

A:模拟辗转相除法时记录答案 B:3种情况:能降低2,能降低1.不能降低分别考虑清楚 C:利用一个set和一个multiset,把行列分开考虑.利用set自带的排序和查询.每次把对应的块拿出来分成两块插入回去.然后行列分别取最大相乘的作为这次询问的答案 D:一个区间覆盖问题的变形.注意公式的话.非常easy发现事实上x.w相应的就是一个[x - w, x + w]的区间,然后求最多不重合区间就可以 代码: #include <cstdio> #include <cstring> #…

题意问,这个点的然后求子树的第i个节点. 这道题是个非常明显的DFS序: 我们只需要记录DFS的入DFS的时间,以及出DFS的时间,也就是DFS序, 然后判断第i个子树是否在这个节点的时间段之间. 最后用一个映射,把相应DFS序对应的节点编号写入.即可 #include<iostream> #include<stdio.h> #include<string.h> #include<algorithm> #include<vector> using…

A. Glass Carving time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a re…

B. Error Correct System time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Ford Prefect got a job as a web developer for a small company that makes towels. His current work task is to create…

传送门 C. Glass Carving time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has…

题目大意: 一开始第一行是 1,第二行是2 4 ,第三行是3 5 7 9 ,类似这样下去,每一行的个数是上一行的个数,然后对这些点从第一个进行编号,问你从[l,r]区间数的和. 思路:分别求出奇数和偶数的个数.然后开始暴力.居然过了== 公式;前m个奇数的和是(m^2),前m个偶数的和是(m*(m+1)). #include<bits/stdc++.h> using namespace std; #define int unsigned long long #define mod 100000…

Ford Prefect got a job as a web developer for a small company that makes towels. His current work task is to create a search engine for the website of the company. During the development process, he needs to write a subroutine for comparing strings S…

题意:建一颗以\(1\)为根结点的树,询问\(q\)次,每次询问一个结点,问该结点的第\(k\)个子结点,如果不存在则输出\(-1\). 题解:该题数据范围较大,需要采用dfs预处理的方法,我们从结点\(1\)开始向下找,\(ans\)数组记录的是,第\(x\)次查找时的结点,\(path\)表示某个结点所需的查找次数,\(siz\)数组表示某个结点的子结点个数.之后每次询问时,判断一下情况是否成立,如果成立,先找出该结点所对应的查找次数(\(path[u]\)),向下找第\(k\)个子结点即可…