B. Interesting drink】的更多相关文章

Interesting drink 题目链接: http://codeforces.com/contest/706/problem/B Description Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bough…
题目链接: B. Interesting drink 题意: 给出第i个商店的价钱为x[i],现在询问mi能在多少个地方买酒; 思路: sort后再二分; AC代码: #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <map> #include <bits/stdc++.…
Interesting drink Problem Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in n different shops in the city. It's known that th…
题意:给定 n 个数,然后有 m 个询问,每个询问一个数,问你小于等于这个数的数有多少个. 析:其实很简单么,先排序,然后十分查找,so easy. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <…
题目链接: http://codeforces.com/problemset/problem/706/B 题目大意: n (1 ≤ n ≤ 100 000)个商店卖一个东西,每个商店的价格Ai,你有m(1≤m≤100 000)天,每天有Cj的钱,问每天可以负的起的商店数. 题目思路: [二分] 排个序,二分. // //by coolxxx // #include<iostream> #include<algorithm> #include<string> #inclu…
排序,二分. 将$x$数组从小到大排序,每次询问的时候只要二分一下位置就可以了. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<vector> #include<map> #include&l…
链接 [http://codeforces.com/group/1EzrFFyOc0/contest/706/problem/B] 题意 给你n个数,q次查询,每次输入一个m,问n个数中有多少个数小于等于m 思路 先排序,再用upper_bound找,该函数返回第一个大于m的下标. 代码: #include<bits/stdc++.h> using namespace std; #define ll long long int main(){ ios::sync_with_stdio(fals…
https://codeforces.com/problemset/problem/706/B 因为没有看见 $x_i$ 的上限是 $10^5$ ,就用了二分去做,实际上这道题因为可乐的价格上限是 $10^6$ ,可以用复杂度为 $O(max(x_i))$ 的dp去做. 也就是说,当这道题的可乐数量上升,二分就容易超时,而可乐的价格上升则dp容易爆内存且超时.各有所长 #include<bits/stdc++.h> using namespace std; #define ll long lo…
A题 Beru-taxi 随便搞搞.. #include <cstdio> #include <cmath> using namespace std; int a,b,n; struct _ { int x,y,v; }p[]; double dis2(_ A) { return (a - A.x) * (a - A.x) + (b - A.y) * (b - A.y); } int main() { scanf("%d%d%d", &a, &b…
1.CF 706B  Interesting drink 2.链接:http://codeforces.com/problemset/problem/706/B 3.总结:二分 题意:给出n个数,再给出q个mi,每次求n个数里有多少个数<=mi #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm> #include&l…