Couple Cover Time Limit: 3000MS   Memory Limit: 524288KB   64bit IO Format: %I64d & %I64u Description 方宝宝有n个篮球,每个篮球上写有一个值ai.他第一次从n个篮球中取出1个,不放回.第二次再在剩余的篮球中取出一个. (每个球被取概率相同).如果这两个球上的值的乘积大于等于p,他会变得高兴,然后请大家吃饭.否则方宝宝会不高兴, 然后暴食暴饮变得更胖. 当然为了方宝宝的健康(被请吃饭),我想取一个…

分析:开一个300w的数组,统计,然后nlogn统计每个值在在序对第一个出现有多少种情况 时间复杂度:O(nlogn) n在3e6数量级 #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> using namespace std; typedef long long LL;…

可以暴力预处理出每一种小于3000000的乘积有几种.询问的时候可以用总的方案减去比p小的有几种o(1)输出. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cmath> #include<cstdlib> #include<algorithm> #include<ve…

题目链接: F. Couple Cover time limit per test 3 seconds memory limit per test 512 megabytes input standard input output standard output Couple Cover, a wildly popular luck-based game, is about to begin! Two players must work together to construct a recta…

传送门:https://codeforces.com/contest/691/problem/F 题意:给你n个数和q次询问,每次询问问你有多少对ai,aj满足ai*aj>=q[i],注意 a*b 与 b*a是不同的 题解:sum[i]记录的是两个数乘积为i的方法数,然后前缀和记录小于等于乘积为i的方法个数,输出答案就容斥一下,因为n个数最多组成n*(n-1)/2对数,减去小于乘积为q[i]的数后即为乘积大于等于q[i]的方法个数 为什么可以暴力是因为注意到了值域的范围为3e6,调和级数的复杂度…

Couple Cover, a wildly popular luck-based game, is about to begin! Two players must work together to construct a rectangle. A bag with nballs, each with an integer written on it, is placed on the table. The first player reaches in and grabs a ball ra…

题目链接: B. Restoring Painting time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he…

D. Jerry's Protest time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output Andrew and Jerry are playing a game with Harry as the scorekeeper. The game consists of three rounds. In each round, Andr…

题目链接: https://vjudge.net/problem/1735275/origin 基本思路: 本题思路比较简单,首先,我们知道 a & x = b, b & x = b; 所以,一个数通过与运算改变只能改变一次! 所以,这里就有一种暴力的写法,三次for循环,时间复杂度是O(3n). 第一次,从1到n遍历一遍,计算各个元素出现的次数,如果大于等于2,则输出0,结束. 第二次,从1到n遍历,如果存在 vis[ a[i]&k ] >= 2 并且 与k与运算后的值与之…

题目链接: http://codeforces.com/contest/651/problem/D D. Image Preview time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vasya's telephone contains n photos. Photo number 1 is currently opened on…