• Codeforces Round #345 (Div. 2) B


    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.

    题解:在一串序列中  不断的找到单调递增序列 序列中的元素只能用一次

    标记处理

     1 #include<bits/stdc++.h>
     2 #include<iostream>
     3 #include<cstdio>
     4 #define ll __int64
     5 using namespace std;
     6 int a[1005],n,ss[1005];
     7 int main() {
     8     scanf("%d",&n);
     9     for(int i=1;i<=n;i++)
    10     {
    11         scanf("%d",&a[i]);
    12          ss[a[i]]++;
    13     }
    14     int re=0;
    15     sort(a+1,a+n+1);
    16     while(1)
    17     {
    18         int sm=0;
    19         for(int j=1;j<=1000;j++)
    20         {
    21             if(ss[j])
    22             {
    23                 sm++;
    24                 ss[j]--;
    25             }
    26         }
    27         if(!sm)break;
    28         re+=(sm-1);
    29     }
    30    printf("%d
    ",re);
    31     return 0;
    View Code
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  • 原文地址:https://www.cnblogs.com/hsd-/p/5255137.html
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