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 Routine Problem time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Manao has a monitor. The screen of the monitor has horizontal to vertical length ratio a:b. Now he is going to watch a movie…
题目链接:http://codeforces.com/problemset/problem/337/B 看到这个题目,觉得特别有意思,因为有熟悉的图片(看过的一部电影).接着让我很意外的是,在纸上比划了一下,凭着直觉,竟然一次AC,那个兴奋啊 !^_^ !  好啦,不说废话. 这个题目被分类为 math 和 matrices ,数学还好理解,matrices,应该是母函数吧(不好意思的说,还没系统地学到),姑且让我分类到数学里吧.题目的意思是,给出一个水平长度 : 垂直长度的比例分别为 a:b…
UVa10025 ? 1 ? 2 ? ... ? n = k problem The problem Given the following formula, one can set operators '+' or '-' instead of each '?', in order to obtain a given k? 1 ? 2 ? ... ? n = k For example: to obtain k = 12 , the expression to be used will be:…
题目链接: POJ:id=3100" style="font-size:18px">http://poj.org/problem? id=3100 ZOJ:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1818 HDU:pid=2740">http://acm.hdu.edu.cn/showproblem.php?pid=2740 Description Given positive…
Kevin's Problem 题目连接: https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4921 Description In his applied probability class, Kevin learned about the Secretary Problem. There are N applicants a…
pid=5105" target="_blank" style="">题目链接:hdu 5105 Math Problem 题目大意:给定a.b,c,d.l,r.表示有一个函数f(x)=|a∗x3+b∗x2+c∗x+d|(L≤x≤R),求函数最大值. 解题思路:考虑极点就可以,将函数求导后得到f′(x)=0的x,即为极值点.在极值点处函数的单调性会发生变化,所以最大值一定就在区间边界和极值点上.注意a=0.b=0的情况,以及极值点不在区间上. #in…
题目链接: Hard problem Time Limit: 2000/1000 MS (Java/Others)     Memory Limit: 65536/65536 K (Java/Others) Problem Description cjj is fun with math problem. One day he found a Olympic Mathematics problem for primary school students. It is too difficult…
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…
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…
官方题解: f(x)=|a∗x3+b∗x2+c∗x+d|, 求最大值.令g(x)=a∗x3+b∗x2+c∗x+d,f(x)的最大值即为g(x)的正最大值,或者是负最小值.a!=0时, g′(x)=3∗a∗x2+2∗b∗x+c 求出g′(x)的根(若存在,x1,x2,由导数的性质知零点处有极值.ans=max(f(xi)|L≤xi≤R).然后考虑两个端点的特殊性有ans=max(ans,f(L),f(R)). 当时 x = -c/(2*b) 写成 x = -c/2*b 了,然后过pretest了.…