【Codeforces Round 345 (Div 2) B】【贪心 STL-set】Beautiful Paintings 安排顺序使得a[i]大于a[i-1]的i尽可能多

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

4200 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.

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 0, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
multiset<int>sot;
multiset<int>::iterator it;
int n, x;
int main()
{
while (~scanf("%d", &n))
{
sot.clear();
while (n--)scanf("%d",&x),sot.insert(x);
int now = *sot.begin();
sot.erase(sot.begin());
int ans = 0;
while(!sot.empty())
{
it = sot.upper_bound(now);
if (it == sot.end())it = sot.begin();
else ++ans;
now = *it;
sot.erase(it);
}
printf("%d\n", ans);
}
return 0;
}
/*
【题意】
有n（1000）个数a[]，数值在[1,1000]
问你如何安排，可以使得这个数组中存在最多的位置i，
满足a[i]>a[i-1]

【类型】
贪心 STL-set

【分析】
显然，我们每次取最小的数。
然后找到比它大的数中尽可能小的那个。
找不到再从当前最小的数开始取。
贪到最后就是答案。

可以用排序+标记数组或set实现

【时间复杂度&&优化】
O(nlogn)

*/