Anton and Making Potions time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Anton is playing a very interesting computer game, but now he is stuck at one of the levels. To pass to the next le…

C. Anton and Making Potions 题目连接: http://codeforces.com/contest/734/problem/C Description Anton is playing a very interesting computer game, but now he is stuck at one of the levels. To pass to the next level he has to prepare n potions. Anton has a…

C. Anton and Making Potions time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Anton is playing a very interesting computer game, but now he is stuck at one of the levels. To pass to the next…

题目链接:http://codeforces.com/contest/734/problem/C C. Anton and Making Potions time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Anton is playing a very interesting computer game, but now he i…

codeforces734C——Anton and Making Potions详解 Anton and Making Potions 题目描述(google翻译) 安东正在玩一个非常有趣的电脑游戏,但现在他被困在其中一个级别.为了进入下一个级别,他必须准备n个药水. 安东有一个特殊的水壶,可以在x秒内准备一个魔药.此外,他知道两种类型的法术可以加快准备魔药的过程. 这种类型的法术加速了一种魔药的准备时间.有这种类型的m个法术,其中第i个成本为bi manooints并且将每个药水的准备时间改为…

CodeForce-734C Anton and Making Potions  C. Anton and Making Potions time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Anton is playing a very interesting computer game, but now he is stuck…

Anton and Making Potions time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Anton is playing a very interesting computer game, but now he is stuck at one of the levels. To pass to the next le…

Anton is playing a very interesting computer game, but now he is stuck at one of the levels. To pass to the next level he has to prepare npotions. Anton has a special kettle, that can prepare one potions in x seconds. Also, he knows spells of two typ…

题目链接:http://codeforces.com/problemset/problem/734/C 题目大意:要制作n个药,初始制作一个药的时间为x,魔力值为s,有两类咒语,第一类周瑜有m种,每种咒语使制作一个药的时间变成a[i],花费b[i]的魔力,第二类咒语有k种,每种咒语瞬间产生c[i]个药,花费d[i]的魔力,c[i]和d[i]都是不递减的,求最短时间内产生n个药的时间.解题思路:因为c[i]和d[i]都是不降的,所以可以枚举a[i],然后二分查找花费小于t-b[i]的第二类咒语.注…

http://codeforces.com/contest/734/problem/C 因为有两种操作,那么可以这样考虑, 1.都不执行,就是开始的答案是n * x 2.先执行第一个操作,然后就会得到一个time和left.就是你会得到一个新的用时,和一个剩下的魔法数,然后在第二个操作数中二分,二分第一个小于等于left的值,意思就是我现在还拥有left点魔法,能够买最多多少个技能的意思. 就是,看着样例一 得到的会是 time : 40s    80s 60s left   : 79 89 5…