题目链接:Codeforces 484B Maximum Value 题目大意:给定一个序列,找到连个数ai和aj,ai%aj尽量大,而且ai≥aj 解题思路:类似于素数筛选法的方式,每次枚举aj,然后枚举k,每次用二分找到小于k∗aj而且最大的ai,维护答案,过程中加了一些剪枝. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn =…
题目链接: B. Skills time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Lesha plays the recently published new version of the legendary game hacknet. In this version character skill mechanism was…
B. Tennis Game time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Petya and Gena love playing table tennis. A single match is played according to the following rules: a match consists of mult…
题目描述: Maximum Value time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given a sequence a consisting of n integers. Find the maximum possible value of (integer remainder of *a**i* divi…
Yukari's Birthday  HDU4430 就是枚举+二分: 注意处理怎样判断溢出...(因为题目只要10^12) 先前还以为要用到快速幂和等比数列的快速求和(但肯定会超__int64) 而且这样判断会超时的... 还有题目中的And it's optional to place at most one candle at the center of the cake. (中间的蜡烛可有可无) 还有观察数据就知道:因为n最大10^12,r最多枚举到40,然后二分k的结果,看是否有符合条…
1514: Packs Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 61  Solved: 4[Submit][Status][Web Board] Description Give you n packs, each of it has a value v and a weight w. Now you should find some packs, and the total of these value is max, total of…
The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute how many quadruplet (a, b, c, d ) ∈ A x B x C x D are such that a + b + c + d = 0 . In the following, we assume that all lists have the same size n .…
You are given an array a consisting of n integers, and additionally an integer m. You have to choose some sequence of indices b1, b2, ..., bk (1 ≤ b1 < b2 < ... < bk ≤ n) in such a way that the value of is maximized. Chosen sequence can be empty.…
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…
题目链接: http://codeforces.com/problemset/problem/484/B 题意: 求a[i]%a[j] (a[i]>a[j])的余数的最大值 分析: 要求余数的最大值非常明显a[i]越接近a[j]的倍数则余数越大 ,因此我们将全部的元素从大到小排序 : 然后枚举a[j]的倍数 ,二分查找小于a[i]倍数的最大值,然后更新余数的最大值. 代码例如以下: #include <iostream> #include <cstdio> #include…
E. Simple Skewness time limit per test:3 seconds memory limit per test:256 megabytes input:standard input output:standard output Define the simple skewness of a collection of numbers to be the collection's mean minus its median. You are given a list…
传送:http://codeforces.com/gym/101612 题意:给定一个数n(<=1e18),将n分解为若干个数的成绩.要求这些数两两之间的差值不能大于1. 分析: 若n==2^k,则答案一定是-1. 然后,考虑若n==a^k,枚举k,二分求a.若n==a^x*(a+1)^y,枚举x,y,二分求解a. 注意:两数相乘可能>1e18,特判. #include<bits/stdc++.h> using namespace std; typedef long long ll…
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…
A. Sagheer and Crossroads time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two…
题目描述: Prime Gift time limit per test 3.5 seconds memory limit per test 256 megabytes input standard input output standard output Opposite to Grisha's nice behavior, Oleg, though he has an entire year at his disposal, didn't manage to learn how to sol…
https://codeforces.com/contest/1241/problem/C You are an environmental activist at heart but the reality is harsh and you are just a cashier in a cinema. But you can still do something! You have n tickets to sell. The price of the i-th ticket is pi.…
888E - Maximum Subsequence 思路:折半枚举. 代码: #include<bits/stdc++.h> using namespace std; #define ll long long #define pb push_back #define mem(a,b) memset(a,b,sizeof(a)) ; int a[N]; set<int>s; int main() { ios::sync_with_stdio(false); cin.tie(); i…
题意: 若干张照片,从头开始可以向左右两边读,已经读过的不需要再读,有的照片需要翻转,给定读.滑动和翻转消耗的时间,求在给定时间内最多能读多少页? 分析: 首先明确,只横跨一次,即先一直读一边然后再一直读另一边,这样消耗的滑动时间最少.是否能在给定时间内读完页数很好判断,所以用二分+枚举,先枚举左边的所有可能情况,再二分右边求出最大页数, 再枚举右边,求出左边.取两边的最大值即可. 代码: #include<iostream> #include<cstdio> #include&l…
C - Preparing for the Contest Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 377B Description Soon there will be held the world's largest programming contest, but the testing system sti…
原题传送门 E. Maximum Subsequence time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given an array a consisting of n integers, and additionally an integer m. You have to choose some sequen…
题目描述: Match Points time limit per test 2 seconds memory limit per test 256 mega bytes input standard input output standard output You are given a set of points x1, , ..., *x**n* Two points iand jcan be matched with each other if the following conditi…
Vasya and Beautiful Arrays CodeForces - 354C Vasya's got a birthday coming up and his mom decided to give him an array of positive integers a of length n. Vasya thinks that an array's beauty is the greatest common divisor of all its elements. His mom…
原题链接:http://codeforces.com/contest/830/problem/A 题意:在一条数轴上分别有n个人和k把钥匙(n<=k),以及一个目的地,每个人要各自拿到一个钥匙后到达目的地.每个人的移动速度都是1, 问所有人都到达目的地的最短时间. 思路:转化一下题意,就是求耗时最长的人所用的最短时间. 我们可以二分答案x,然后对排序后的人以及钥匙进行枚举,进行从左至右搭配. 这里check函数中返回false的条件是从左至右所有人都能在x的时间内到达目的地,而计算这些人到达目的…
题目:Maximum Matching 传送门:http://codeforces.com/contest/1038/problem/E 分析: 一个块拥有{color1,val,color2},两个块相连要求相连处颜色相同,求价值最大的连接方案. 关心到color最大为4,以4种颜色为点,对于每个块,在(color1,color2)间连一条边权为(val)的边,建一张4个点n条边的图.显然,在图上选一条价值最大的路径(或回路)就是答案了. 方法一: 如果这张图本身就是Eular路径(或Eula…
链接:https://codeforces.com/contest/1288/problem/D D. Minimax Problem 题意:给定n个数组,长度为m,从n中数组挑选两个数组,两个数组中的每一位取两者的最大值组成一个新的数组,新数组中的最小值记为c,所有组合中c的最大值 思路:题目中m的范围只有8,数组中元素的范围是1e9,显然m的范围非常特殊,似乎可以把数组用二进制的形式进行状态压缩,这里采用二分答案的方法.二分范围是数组元素最小值到最大值,每次check(mid)一遍,chec…
E - E CodeForces - 1100E 一个n个节点的有向图,节点标号从1到n,存在m条单向边.每条单向边有一个权值,代表翻转其方向所需的代价.求使图变成无环图,其中翻转的最大边权值最小的方案,以及该方案翻转的最大的边权. Input 单组输入,第一行包含两个整数n和m(2≤n≤100 000,1≤m≤100 000) 接下来m行,每行3个整数,u_i ,v_i ,w_i (1<= u_i , v_i <= n, 1<= w_i <= 10^9),表示u到v有一条权值为w…
题意:问方程X^Z + Y^Z + XYZ = K (X<Y,Z>1)有多少个正整数解 (K<2^31) 解法:看K不大,而且不难看出 Z<=30, X<=sqrt(K), 可以枚举X和Z,然后二分找Y,这样的话不把pow函数用数组存起来的话好像会T,可以先预处理出1~47000的2~30次幂,这样就不会T了. 但是还可以简化,当Z=2时,X^2+Y^2+2XY = (X+Y)^2 = K, 可以特判下Z= 2的情况,即判断K是否为平方数,然后Z就可以从3开始了,这样的话X^…
题目链接:http://poj.org/problem?id=3977 给你n个数,找到一个子集,使得这个子集的和的绝对值是最小的,如果有多种情况,输出子集个数最少的: n<=35,|a[i]|<=10e15 子集个数共有2^n个,所以不能全部枚举,但是可以分为两部分枚举: 枚举一半的所有情况,然后后一半二分即可: #include<iostream> #include<algorithm> #include<string.h> #include<st…
Problem Description OO has got a array A of size n ,defined a function f(l,r) represent the number of i (l<=i<=r) , that there's no j(l<=j<=r,j<>i) satisfy ai mod aj=0,now OO want to know ∑i=1n∑j=inf(i,j) mod (109+7).   Input There are m…
Delay Constrained Maximum Capacity Path Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1839 Description Consider an undirected graph with N vertices, numbered from 1 to N, and M edges. The vertex numbered with…