Alice is providing print service, while the pricing doesn't seem to be reasonable, so people using her print service found some tricks to save money.For example, the price when printing less than 100 pages is 20 cents per page, but when printing not…

Alice's Print Service Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1855    Accepted Submission(s): 454 Problem Description Alice is providing print service, while the pricing doesn't seem to…

Alice's Print Service Time Limit: 2 Seconds      Memory Limit: 65536 KB Alice is providing print service, while the pricing doesn't seem to be reasonable, so people using her print service found some tricks to save money. For example, the price when…

A - Alice's Print Service Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4791 Description Alice is providing print service, while the pricing doesn't seem to be reasonable, so people using her…

Alice's Print Service Time Limit: 2 Seconds      Memory Limit: 65536 KB Alice is providing print service, while the pricing doesn't seem to be reasonable, so people using her print service found some tricks to save money. For example, the price when…

Alice is providing print service, while the pricing doesn't seem to be reasonable, so people using her print service found some tricks to save money. For example, the price when printing less than 100 pages is 20 cents per page, but when printing not…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4791 解题报告:打印店提供打印纸张服务,需要收取费用,输入格式是s1 p1 s2 p2 s3 p3...表示打印区间s1到s2张纸的单价是p1,打印区间s2 到s3的单价是p2....最后是sn到无穷大的单价是pn,让你求打印k张纸的总费用最少是多少?有m次查询. 因为s1*p1 > s2 * p2 > s3*p3......,很显然,加入k所在的那个区间是第x个区间,那么最低费用要么是k * p…

题意:就是一个打印分段收费政策,印的越多,单张价格越低,输入需要印刷的数量,求最小印刷费用一个细节就是,比当前还小的状态可能是最后几个. #include<stdio.h> #include<string.h> #include<cstdio> #include<string> #include<math.h> #include<algorithm> #include<iostream> #define LL long l…

点击打开链接 Alice's Print Service Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1084    Accepted Submission(s): 240 Problem Description Alice is providing print service, while the pricing doesn't…

题目链接: HDU:http://acm.hdu.edu.cn/showproblem.php?pid=4791 ZJU:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5072 Problem Description Alice is providing print service, while the pricing doesn't seem to be reasonable, so people using her pr…

POJ 2296 Map Labeler / ZOJ 2493 Map Labeler / HIT 2369 Map Labeler / UVAlive 2973 Map Labeler(2-sat 二分) Description Map generation is a difficult task in cartography. A vital part of such task is automatic labeling of the cities in a map; where for e…

/** 题目: uvalive 3231 Fair Share 公平分配问题 链接:https://vjudge.net/problem/UVALive-3231 题意:有m个任务,n个处理器,每个任务有两个候选处理器,只要其中一个运行,该任务就能执行. 不同任务的两个候选处理器,至少有一个不同. 求任务数最多的那个处理器所分配的任务数尽量少. 思路:二分+最大流 左边是任务,s->u,cap = 1. 如果任务u和u的候选处理器v,u->v, cap = 1. 右边是处理器,二分mi.所有处…

题目链接:https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=5273 Wei Qing (died BC) was a military general of the Western Han dynasty whose campaigns against the Xiongnu earned him great acclaim.…

题目链接: http://vjudge.net/problem/viewProblem.action?id=25636 对于>1的堆,必然会被其中一人全部合并. 然后就是二维dp,dp[非1堆的操作数][1堆个数]. #include <stdio.h> #include <cstring> #include <cstdlib> #include <algorithm> #include <cmath> using namespace st…

题意:有 n 堆石子,有两种操作,一种是从一堆中拿走一个,另一种是把两堆合并起来,Alice 先拿,谁不能拿了谁输,问谁胜. 析:某些堆石子数量为 1 是特殊,石子数量大于 1 个的都合并起来,再拿,这是最优的,因为都想另一个输,并且第二种操作是可以翻转胜负的,所以都会先采取第二个操作,但是砘数量为 1 却不是,所以要分开考虑,dp[i][j] 表示,数量为 1 的堆的个数,总的操作数为 j,先手胜还是负. 考虑边界,如果剩下的都是 1 的,那么 i % 3 != 0 先手胜, 如果没有 1了,…

全场最水题. 保留打印a[i]份分别需要的钱,从后往前扫一遍,保证取得最优解. 查找的时候,二分同时判断最小值即可. 注意初值的设定应该设定为long long 的无穷大. #include <iostream> #include <cstdio> #include <cstring> #define maxn 100100 typedef long long ll; using namespace std; ll a[maxn],b[maxn],c[maxn],f[m…

题意是黑板上有n个数\(S_i\).每次操作可以把其中一个数减1或者将两个数合并为一个数.一个数变为0时,则不能再对其操作. 思路是发现最大的可操作次数为\( \sum S_i\)+(n - 1).\( \sum S_i\)是把所有数消除需要的操作数.(n-1)表示我们最多可以合并(n-1)次. 同时我们发现,总操作数的奇偶决定了胜负.换言之,合并的次数决定论胜负. 当1个数为1时,假如我们将其消去,则合并的次数减1,总操作数的奇偶改变. 那么我们首先考虑,所有的数都>=2的情况,假如这种情况对…

分析: 1.由于价格是递减的,所以可能出现si*pi>sj*pj(j>i).所以要有一个数组来储存当前端点的最小值. 2.然后二分查找当前的si,比较q*p[i]和M[i+1].不过在这之前要确认i是小于n的.] 3.upper_bound是返回第一个大于当前值得坐标,否则返回左闭右开的右端点.而lower_bound是返回第一个大于等于当前值得坐标.所以这里采用upper_bound. #include <iostream> #include <cstdio> #in…

题意:找一个出现了m次的最长子串,以及这时的最右的位置. hash的话代码还是比较好写的,,但是时间比SA多很多.. #include <stdio.h> #include <algorithm> #include <string.h> using namespace std; + ; typedef long long ll; ; ; char s[N]; int m,len,pw[N]; int H[N],pos; struct node { int id,hash…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4791 题目解释:给你一组数据s1,p1,s2,p2...sn,pn,一个数字q,问你当要打印q张资料时,最少花费是多少?值得注意的是p1>p2>p3>...>pn,就是因为看清这个条件,走了很多弯路. 代码: #include<cstdio> #include<cstring> #include<cstdlib> #include<algori…

Alice's Print Service Time Limit: 2 Seconds      Memory Limit: 65536 KB Alice is providing print service, while the pricing doesn't seem to be reasonable, so people using her print service found some tricks to save money. For example, the price when…

public class MyService extends Service { public MyService() { } @Override public IBinder onBind(Intent intent) { // TODO: Return the communication channel to the service. //throw new UnsupportedOperationException("Not yet implemented"); return n…

Server部分: #!/usr/bin/env python import sys import os import rospy #from beginner.srv import * from beginner.srv import AddTwoInts def add_two_ints_client(x,y): rospy.wait_for_service('add_two_ints')# rospy.wait_for_service(‘service的tipoc’) try: add_t…

Server部分: #!/usr/bin/env python import sys import os import rospy #from beginner.srv import * from beginner.srv import AddTwoInts def add_two_ints_client(x,y): rospy.wait_for_service('add_two_ints')# rospy.wait_for_service(‘service的tipoc’) try: add_t…

import bisectL = [1, 3, 3, 6, 8, 12, 15]x = 5x_insert_point = bisect.bisect_left(L, x)# 在L中查找x,x存在时返回x左侧的位置,x不存在返回应该插入的位置..这是3存在于列表中,返回左侧位置1print(x_insert_point)x_insert_point = bisect.bisect_right(L, x) # 在L中查找x,x存在时返回x右侧的位置,x不存在返回应该插入的位置..这是3存在于列表中…

Service跟Windows系统里的服务概念差不多,都在后台执行,它跟Activity的最大区别就是,它是无界面的 开发Service与开发Activity的步骤类似 1.定义一个继承Service的子类 2.在AndroidManifest.xml文件中配置该Service Service与Activity还有一点相似之处,它们都是从Context派生出来的,因此它们都可调用Context里定义的如getResources(),getContentResolver()等方法 下面写一个简单的…

time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output   It is so boring in the summer holiday, isn't it? So Alice and Bob have invented a new game to play. The rules are as follows. First, they…

A method implemented by a network device residing in a service domain, wherein the network device comprises an information centric networking (ICN) transport layer and a service access layer (SAL) for handling context-aware service logistics and serv…

插入排序 算法分析 两次循环, 大循环对队列中的每一个元素拿出来作为小循环的裁定对象 小循环对堆当前循环对象在有序队列中寻找插入的位置 性能参数 空间复杂度 O(1) 时间复杂度 O(n^2) 详细代码解读 import random def func(l): # 外层循环: 对应遍历所有的无序数据 for i in range(1, len(l)): # 备份 取出数据 temp = l[i] # 记录取出来的下标值 pos = i # 内层循环: 对应从后往前扫描所有有序数据 ""…

It is so boring in the summer holiday, isn't it? So Alice and Bob have invented a new game to play. The rules are as follows. First, they get a set of n distinct integers. And then they take turns to make the following moves. During each move, either…