最近刚学完数值分析上的方程求根——牛顿法,所以做几题练习一下。 Problem Description Now, here is a fuction: F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100) Can you find the minimum value when x is between 0 and 100.
This way 题意: 二维平面上有一些点,你现在有一个没有顶边的矩形,问你有多少种包含点的情况(每个点视为不同) 题解: 将每个点视为矩形下底边上的点,查找这个点左边有多少点,右边有多少点,这个点做完之后将其删除,相同高度的点从左到右做,对于右边的点要注意左端点位左边的点+1: 这张图就表示了相同高度右边点的可查询区间。 (刚多校结束发现2200真的是比赛中的简单题了) #incl
原题: Raju has recently passed BSc. Engineering in Computer Science & Engineering from BUET (Bangladesh University of Extraordinary Talents), the best university of Bangladesh. After passing, he has be
G - Vitya and Strange Lesson CodeForces - 842D Today at the lesson Vitya learned a very interesting function — mex. Mex of a sequence of numbers is the minimum non-negative number that is not pres
题目: There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 <= Ki <= N) on every floor.The lift have just two buttons: up and down.When you at floor i,if
题意 输出n个盘子在4个塔的汉诺塔问题最少要多少步。 思路 我们设f[n]为n个盘子在4塔的汉诺塔问题下需要的最少步数,d[i]为i个盘子在3塔的汉诺塔问题下需要的最少步数,可以得出动态转移方程: f[n]=min(2∗f[i]+d[n−i]) f [ n ] = m i n ( 2 ∗ f [ i ] + d [ n − i ] ) f[n]=min(2*f[i]+d[n-i])
题意: 在一个 n x n 的平面上,给定 m 个等腰直角三角形(各点均为整数),问该平面上被三角形覆盖奇数次的点有多少个。 思路: 由于 n 较大,不能模拟解决,故使用离散化思想。 考虑每一行有多少点被覆盖了奇数次,题目从二维转换成一维。 对于每一行,考虑每个三角形在此行覆盖的线段,记录下每条线段的左端点 l 、右端点 r 保存在同一个数组中。 排序后则容易知道第一个到第二个数、第三到第四个