“玲珑杯”ACM比赛 Round #5 B -- Private Value【STL-map】

B -- Private Value

Time Limit：1s Memory Limit：128MByte

Submissions：526Solved：174

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

SOLUTION

玲珑杯”ACM比赛 Round #5

原题链接：http://www.ifrog.cc/acm/problem/1057

题意：输出一个序列中出现次数最少的元素，数组可能会超时，类似于桶排序的方法估计也会超时，我用map就不用考虑那么多了。

AC代码：

/**
* 行有余力，则来刷题！
* 博客链接:http://blog.csdn.net/hurmishine
*
*/
#include <iostream>
#include <map>
using namespace std;
int main()
{
int T;
cin>>T;
while(T--)
{
int n;
cin>>n;
map<int,int>mp;
mp.clear();
int x;

for(int i=0; i<n; i++)
{
cin>>x;
mp[x]++;
}
map<int,int>::iterator iter;
int minn=1e9;
for(iter=mp.begin(); iter!=mp.end(); iter++)
{
if(iter->second<minn)
minn=iter->second;
}
bool first=true;
for(iter=mp.begin(); iter!=mp.end(); iter++)
{
if(iter->second==minn)
{
if(!first)
cout<<" "<<iter->first;
else
{
first=false;
cout<<iter->first;
}
}
}
cout<<endl;
}
return 0;
}