zoukankan      html  css  js  c++  java
  • Codeforces Round #345 (Div. 2)——B. Beautiful Paintings(贪心求上升序列个数)

    B. Beautiful Paintings
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    There are n pictures delivered for the new exhibition. The i-th painting has beauty ai. We know that a visitor becomes happy every time he passes from a painting to a more beautiful one.

    We are allowed to arranged pictures in any order. What is the maximum possible number of times the visitor may become happy while passing all pictures from first to last? In other words, we are allowed to rearrange elements of a in any order. What is the maximum possible number of indices i (1 ≤ i ≤ n - 1), such that ai + 1 > ai.

    Input

    The first line of the input contains integer n (1 ≤ n ≤ 1000) — the number of painting.

    The second line contains the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000), where ai means the beauty of the i-th painting.

    Output

    Print one integer — the maximum possible number of neighbouring pairs, such that ai + 1 > ai, after the optimal rearrangement.

    Examples
    input
    5
    20 30 10 50 40
    
    output
    4
    
    input
    4
    200 100 100 200
    
    output
    2
    
    Note

    In the first sample, the optimal order is: 10, 20, 30, 40, 50.

    In the second sample, the optimal order is: 100, 200, 100, 200.

    题意:求最多的上升序列的个数,用map比数组快多,做法:切蛋糕一样每次切一层 

    代码:

    #include<iostream>
    #include<algorithm>
    #include<vector>
    #include<sstream>
    #include<map>
    using namespace std;
    int main(void)
    {
    	int t,n,i,j,be,sum,ans;
    	while (cin>>n)
    	{
    		map<int,int>list;
    		for (i=0; i<n; i++)
    		{
    			cin>>be;
    			list[be]++;
    		}
    		ans=0;
    		while (n)//每次
    		{
    			int num=0;
    			for (map<int,int>::iterator it=list.begin(); it!=list.end(); it++)
    			{
    				if(it->second>0)
    				{
    					(it->second)--;
    					n--;
    					num++;
    				}
    			}
    			ans=ans+num-1;//两个以上才算一个上升序列
    		}
    		cout<<ans<<endl;
    	}
    	return 0;
    }
  • 相关阅读:
    移动端前端开发调试
    webkit图片滤镜
    ruby安装sass报错解决办法
    mongodb的查询语句学习摘要
    signedCookies
    [cookie篇]从cookie-parser中间件说起
    node.js下mongoose简单操作实例
    在ExpressJS中设置二级域名跨域共享Cookie
    Node.js开发工具、开发包、框架等总结
    hibernate框架学习笔记4:主键生成策略、对象状态
  • 原文地址:https://www.cnblogs.com/Blackops/p/5356409.html
Copyright © 2011-2022 走看看