CodeForces 651B Beautiful Paintings
CodeForces 651B Beautiful Paintings
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 40output
4input
4
200 100 100 200output
2
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
using namespace std;
int a[1005];
int b[1005];
int tag[1005];
int tag2[1005];
int num[1005];
int n;
int main()
{
scanf("%d",&n);
memset(tag2,0,sizeof(tag2));
int cnt=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
if(!tag2[a[i]])
{
tag2[a[i]]=1;
b[++cnt]=a[i];
}
}
sort(b+1,b+cnt+1);
for(int i=1;i<=cnt;i++)
tag[b[i]]=i;
memset(num,0,sizeof(num));
for(int i=1;i<=n;i++)
num[tag[a[i]]]++;
int ans=0;
for(int i=1;i<=cnt;i++)
{
for(int j=i-1;j>=1;j--)
{
if(num[j]<=num[i])
{
ans+=num[j];
num[j]=0;
num[i]-=num[j];
}
else
{
ans+=num[i];
num[i]=0;
num[j]-=num[i];
}
}
}
printf("%d\n",ans);
return 0;
}腾讯云开发者
