zoukankan      html  css  js  c++  java
  • A. Reorder the Array

    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 integers.

    For instance, if we are given an array [10,20,30,40][10,20,30,40], we can permute it so that it becomes [20,40,10,30][20,40,10,30]. Then on the first and the second positions the integers became larger (20>1020>10, 40>2040>20) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals 22. Read the note for the first example, there is one more demonstrative test case.

    Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.

    Input

    The first line contains a single integer nn (1n1051≤n≤105) — the length of the array.

    The second line contains nn integers a1,a2,,ana1,a2,…,an (1ai1091≤ai≤109) — the elements of the array.

    Output

    Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.

    Examples
    input
    Copy
    7
    10 1 1 1 5 5 3
    output
    Copy
    4
    input
    Copy
    5
    1 1 1 1 1
    output
    Copy
    0
    Note

    In the first sample, one of the best permutations is [1,5,5,3,10,1,1][1,5,5,3,10,1,1]. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.

    In the second sample, there is no way to increase any element with a permutation, so the answer is 0.

    给你一个序列,你可以生成这个序列的任意一个排列,对于某个排列,

    如果这个排列上某个位置的值大于原序列的值,那么就会产生1的贡献,问你最大能有多少贡献。

     1 #include<stdio.h>
     2 #include<iostream>
     3 #include<algorithm>
     4 #include<string.h>
     5 #include<map>
     6 using namespace std;
     7 const int maxn = int(1e5) + 7;
     8 int a[maxn], n, r = 1, ans;
     9 int main()
    10 {
    11     cin >> n;
    12     for (int i = 1; i <= n; i++)
    13         cin >> a[i];
    14     sort(a+ 1, a+ 1 + n);
    15     for (int i = 1; i <= n; i++)
    16     {
    17         while (r <= n && a[r] <= a[i])
    18             r++;
    19         if (r <= n)
    20             ans++, r++;
    21     }
    22     cout << ans << std::endl;
    23     return 0;
    24 }
    View Code
  • 相关阅读:
    mac creact-react-app 时包 gyp-error
    JAVA,JDBC报错Access denied for user '"root"'@'localhost
    JAVA,JDBC连接数据库报错:No suitable driver found for "jdbc:mysql://localhost:3306
    JAVA,反射使用demo,通过读取配置文件创建类,调用方法
    JAVA,反射获取class对象的3种方式
    python获取当前时间距离指定时间相差的秒值
    JAVA,字节流文件读写
    JAVA,反射常用方法
    JAVA,字符流读写,flush和close的区别
    JAVA,字符流文件读写
  • 原文地址:https://www.cnblogs.com/luhongkai/p/9517454.html
Copyright © 2011-2022 走看看