玲珑杯-1057-Private Value【思维】【map】

本文介绍了一种算法，用于找出给定整数数组中出现次数最少的元素，并按升序输出这些元素。通过使用映射记录每个元素的出现频率，然后遍历数组找到最小频数并输出所有匹配此频数的元素。

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1057 - Private Value

Time Limit：1s Memory Limit：128MByte

Submissions：562Solved：192

DESCRIPTION

 The "private value" is the value that occurs, but most infrequent.
 An array of integers is given, and you need print the private values of the array.
 Please notice that, there might be several private values, and you need print them in increasing order.

INPUT

   The first line is an integer T(1 <= T <= 4), indicating the number of test cases.For each case, the first line contains an integer n (n <= 100000).The second line contains n integers a_i (0 <= a_i <= 10^9), 
  

OUTPUT

   For each case, print all private values in increasing order in a line.
  

SAMPLE INPUT

    2
   
    7
   
    1 1 2 2 3 3 3
   
    5
   
    5 3 2 4 1

SAMPLE OUTPUT

1 2

    1 2 3 4 5
   

大意：统计序列中出现次数最少的元素，若存在多个，则从小到大输出结果

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<map>
using namespace std;
int n;
int a[100010];
map<int,int> M;
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		M.clear();
		for(int i=0;i<n;i++)
		{
			scanf("%d",a+i);
			M[a[i]]++;
		}
		int mmin=100000;
		for(int i=0;i<n;i++)
		{
			mmin=min(mmin,M[a[i]]);
		}
		sort(a,a+n);
		int top=unique(a,a+n)-a;
		bool flag=0;
		for(int i=0;i<top;i++)
		{
			if(M[a[i]]==mmin)
			{
				if(flag)
					printf(" ");
				printf("%d",a[i]);
				flag=1;
			}
		}
		puts("");
	}
	return 0;
}