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…

题目链接:http://codeforces.com/problemset/problem/691/F 题目大意:给定n个数,再给m个询问,每个询问给一个p,求n个数中有多少对数的乘积≥p 数据范围:2≤n≤10^6, 1≤ai≤3*10^6,1≤m≤10^6, 1≤p≤3*10^6 解题思路:比赛的时候比较naive的思路是把n中的数字排序去了重之后,对于每个p,最多枚举√p步,就能得到答案.而这个naive的思路是O(p√p)的,结果T了. 后来百思不得其解,去看了官方的解答.感觉是一种很有…

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: There are nn points on the plane, (x1,y1),(x2,y2),…,(xn,yn)(x1,y1),(x2,y2),…,(xn,yn). You need to place an isosceles triangle with two sides on the coordinate axis to cover all points (a point is covered if it lies inside the triangle or on the s…

B. Cover Points time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output There are n points on the plane, (x1,y1),(x2,y2),-,(xn,yn). You need to place an isosceles triangle with two sides on the coordin…

题意: 有\(n\)条线段,区间为\([l_i, r_i]\),每次询问\([x_i, y_i]\),问要被覆盖最少要用多少条线段. 思路: \(f[i][j]\)表示以\(i\)为左端点,用了\(2^j\)条线段,最远到哪里. 然后从大到小贪心即可,类似于倍增找LCA的过程. 代码: #include <bits/stdc++.h> using namespace std; #define N 200010 #define M 500010 #define D 20 int n, q; in…