B. Vasya and Wrestling time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by…

题目链接:http://codeforces.com/problemset/problem/493/B 题目意思:给出 n 个 techniques,每个 technique 的值为 ai. ai > 0 表示把这个分数给第一个wrestler,ai < 0,表示给第二个wrestler.约定 ai != 0. 如果两个人最终的得分不相等,分数多的那个人获胜. 如果两个人最终的得分相等,可以分两种情况讨论: (1)序列中至少有一位不相等,那么字典序大的那个获胜.例如第一个人:1, 4, 5,…

B. Vasya and Wrestling 题目连接: http://codeforces.com/contest/493/problem/B Description Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wi…

B. Vasya and Wrestling time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by…

/* CodeForces - 837E - Vasya's Function [ 数论 ] | Educational Codeforces Round 26 题意: f(a, 0) = 0; f(a, b) = 1 + f(a, b-gcd(a, b)); 求 f(a, b) , a,b <= 1e12 分析: b 每次减 gcd(a, b) 等价于 b/gcd(a,b) 每次减 1 减到什么时候呢,就是 b/gcd(a,b)-k 后 不与 a 互质 可先将 a 质因数分解,b能除就除,不能…

CodeForces.158A Next Round (水模拟) 题意分析 校赛水题的英文版,坑点就是要求为正数. 代码总览 #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <sstream> #include <set> #include <map> #include <queue>…

A. Vasya and Football 题目连接: http://codeforces.com/contest/493/problem/A Description Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player wh…

简单题,却犯了两个错误导致WA了多次. 第一是程序容错性不好,没有考虑到输入数据中可能给实际已经罚下场的人再来牌,这种情况在system测试数据里是有的... 二是chronologically这个词没注意,其实如果输入是按时间顺序的,就直接在线处理就行了,用不着int team1[110][110],team2[110][110]两个数组. #include<iostream> #include<cstdio> #include<cstdlib> #include&l…

题意:给定一个2*n的矩形方格,每个格子有一个权值,从(0,0)开始出发,要求遍历完整个网格(不能重复走一个格子),求最大权值和,(权值和是按照step*w累加,step步数从0开始). 转载: 题解:思维题,如果正向考虑的话很容易把自己绕晕,我们需要反过来想,你会发现其实对于一个2*N的矩阵,你一共只有N个终点(如下图1),如果在认真推敲,你会发现对于这n个终点,从起点到终点的路线都是很有规律的,只有下图2和3两种情况)那么问题就简单了,只需要考虑各种前缀的预处理,之后直接O(n)判断这N个终…

http://codeforces.com/problemset/problem/747/C 题意:有n台机器,q个操作.每次操作从ti时间开始,需要ki台机器,花费di的时间.每次选择机器从小到大开始,如果可以完成任务,那么输出id总和,否则输出-1. 思路:简单的模拟,注意如果不能完成任务,那么ser数组是不能更新的. #include <cstdio> #include <algorithm> #include <iostream> #include <cs…