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You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integer…
You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integer…
Bryce1010模板 http://codeforces.com/contest/1008/problems #include <bits/stdc++.h> using namespace std; #define ll long long const int MAXN=1e5+10; const int INF=0x3f3f3f3f; int a[MAXN]; int main() { int n; cin>>n; for(int i=1;i<=n;i++) { cin…
题目链接:https://cn.vjudge.net/problem/CodeForces-1007A 题意 给个数组,元素的位置可以任意调换 问调换后的元素比此位置上的原元素大的元素个数最大多少 思路 一开始想了半天,最后想出来田忌赛马 田忌赛马经典题,一共5种可能性,详见HDU-1052 Tian Ji -- The Horse Racing 贪心 考虑特殊位置(首尾元素)的讨论 提交过程 AC 代码 #include <cstdio> #include <algorithm>…
To CF 这道题是排序贪心,将原序列排序后统计答案即可. 但是直接统计会超时,因为排序后具有单调性,所以可以进行一点优化,这样,便可以通过此题. 而这道题的优化在于单调性,因为 \(a[i+1]\) 必然大于 \(a[i]\),所以当 \(a[j]\) 无法与 \(a[i]\) 匹配时,也就可以排除掉 \(a[i+1]\),原因是因为具有单调性. 因此,不需要重新开始匹配,直接向下匹配即可. 这样,也就可以通过本题,避免超时. #include<cstdio> #include<ios…
[题目] 输入一个整数数组,调整数组中数字的顺序,使得所有奇数位于数组的前半部分,所有偶数位于数组的后半部分.要求时间复杂度为O(n). [分析] 如果不考虑时间复杂度,最简单的思路应该是从头扫描这个数组,每碰到一个偶数时,拿出这个数字,并把位于这个数字后面的所有数字往前挪动一位.挪完之后在数组的末尾有一个空位,这时把该偶数放入这个空位.由于碰到一个偶数,需要移动O(n)个数字,因此总的时间复杂度是O(n2). 要求的是把奇数放在数组的前半部分,偶数放在数组的后半部分,因此所有的奇数应该位于偶数…
Problem2:实现Singleton模式 题目描述:设计一个类,我们只能生成该类的一个实例 package Problem2; public class SingletonClass { /* * 题目描述:设计一个类,我们只能生成该类的一个实例 */ //volatile:防止指令重排序 private static volatile SingletonClass instance; private SingletonClass() { } public static SingletonC…
A - Turn the Rectangles CodeForces - 1008B There are nn rectangles in a row. You can either turn each rectangle by 9090 degrees or leave it as it is. If you turn a rectangle, its width will be height, and its height will be width. Notice that you can…
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