Testing the CATCHER Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 16131   Accepted: 5924 Description A military contractor for the Department of Defense has just completed a series of preliminary tests for a new defensive missile calle…

Language: Default Testing the CATCHER Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 15207   Accepted: 5595 Description A military contractor for the Department of Defense has just completed a series of preliminary tests for a new defen…

Testing the CATCHER Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 13396   Accepted: 4905 Description A military contractor for the Department of Defense has just completed a series of preliminary tests for a new defensive missile calle…

一.Description A military contractor for the Department of Defense has just completed a series of preliminary tests for a new defensive missile called the CATCHER which is capable of intercepting multiple incoming offensive missiles. The CATCHER is su…

题意:题目太长没看,直接看输入输出猜出是最长下降子序列 用了以前的代码直接a了,做法类似贪心,把最小的顺序数存在数组里面,每次二分更新数组得出最长上升子序列 #include<iostream> #include<cstdio> using namespace std; int main() { int dp[40002],a[40002],n,t,i,low,up,top,mid,max,tmp,k,b[40002],cas=1; while(1) { scanf("%…

POJ 1887Testingthe CATCHER (LIS:最长下降子序列) http://poj.org/problem?id=3903 题意: 给你一个长度为n (n<=200000) 的数字序列, 要你求该序列中的最长(严格)下降子序列的长度. 分析:        读取全部输入, 将原始数组逆向, 然后求最长严格上升子序列就可以. 因为n的规模达到20W, 所以仅仅能用O(nlogn)的算法求.        令g[i]==x表示当前遍历到的长度为i的全部最长上升子序列中的最小序列末…

Testing the CATCHER Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 13968   Accepted: 5146 Description A military contractor for the Department of Defense has just completed a series of preliminary tests for a new defensive missile calle…

题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=68553#problem/B 题目要求: Wavio是一个整数序列,具有以下特性: 1.Wavio序列的长度是奇数, 即 L = 2 * n + 1; 2.Wavio序列前 n+1 个整数是递增序列 3.Wavio序列后 n+1 个整数是递减序列 如示例 1 2 3 4 5 4 3 2 1 10 最长的 Wavio序列 为 1 2 3 4 5 4 3 2 1 ,所以答案为9 题…

对于一个二维信号,比如灰度图像,灰度值的范围是0-255,因此只要根据像素灰度值(0-255)出现的概率,就可以计算出信息熵.    但是,对于一个一维信号,比如说心电信号,数据值的范围并不是确定的,不会是(0-255)这么确定,如果进行域值变换,使其转换到一个整数范围的话,就会丢失数据,请高手指点,怎么计算. 比如数字信号是x(n),n=1~N(1)先用Hist函数对x(n)的赋值范围进行分块,比如赋值范围在0~10的对应第      一块,10~20的第二块,以此类推.这之前需要对x(n)做…

求分数序列前N项和 #include <stdio.h> int main() { int i, n; double numerator, denominator, item, sum, swap; while (scanf("%d", &n) != EOF) { numerator = 2; denominator = 1; item = 0; sum = 0; for (i = 1; i <= n; i++) { item = numerator/deno…