Codeforces Round #345 (Div. 2) B. Beautiful Paintings-优快云博客

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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:给出一个初始的序列,叫你判断这个序列重新排号之后最多有多少个前一个小于后一个的数目.

思路:先统计出每个相同的数各有多少个,然后每次都从1-1000扫一遍,看有多少个不同的数cnt同时每个数的数目减一,每次答案加上cnt-1.(即相当于每次求最长上升子序列)

#include<bits/stdc++.h>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=1100;
int cnt[maxn],a[maxn];

int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        memset(cnt,0,sizeof(cnt));
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            cnt[a[i]]++;
        }
        int ans=0;
        for(int i=1;;i++){
            int count=0;
            for(int j=1;j<=1000;j++){
                if(cnt[j]>=1)
                    count++,cnt[j]--;
            }
            if(count<=1)
                break;
            ans+=count-1;
        }
        printf("%d\n",ans);
    }
    return 0;
}

可以变形为ai<=1e9

//对a[i]<=1e9的时候的代码
#include<bits/stdc++.h>
using namespace std;
const int maxn=1010;
multiset<int>st;
int a[maxn],b[maxn];

int DP(int n){
    b[1]=a[1];
    int len=1;
    for(int i=2;i<=n;i++){
        if(a[i]>b[len])
            len=len+1,b[len]=a[i];
        else{
            int pos=lower_bound(b+1,b+len+1,a[i])-b;
            b[pos]=a[i];
        }
    }
    return len;
}

int main(){
    int n,x;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&x);
        st.insert(x);
    }
    int ans=0;
    multiset<int>::iterator it;
    for(int i=1;i<=n;i++){
        int cnt=0;
        for(it=st.begin();it!=st.end();it++)
            a[++cnt]=(*it);
        if(cnt<=0)
            break;
        int len=DP(cnt);
        if(len<=1)
            break;
        for(int i=1;i<=len;i++){
            it=st.lower_bound(b[i]);
            st.erase(it);
        }
        ans+=(len-1);
    }
    printf("%d\n",ans);
    return 0;
}