一开始用了最原始的方法在读入时就去遍历让每个数的存在次数+1,后来发现会超时,到网上看了博客后的做法是先全部读入然后排序,之后从第二个开始遍历,如果是相同的则存在次数+1,不同时输出并重置存在次数。原理很简单,却是完全不同层次的做法,真是佩服。
// Problem#: 19856
// Submission#: 4951234
// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/
// All Copyright reserved by Informatic Lab of Sun Yat-sen University
#include <iostream>
#include <algorithm>
using namespace std;
bool cmp(long long a,long long b)
{
return a < b;
}
int main()
{
int n;
bool notFirst = false;
while(cin >> n)
{
if(notFirst)
{
cout<<endl;
}
notFirst=1;
long long* num=new long long [n+1];
for(int i=0;i<=n-1;i++)
{
cin >> num[i];
}
sort(num,num+n,cmp);
num[n]=0;
int count=1;
for(int j=1;j<=n;j++)
{
if(num[j]!=num[j-1])
{
cout << num[j-1] << " " << count << endl;
count=1;
}
else
{
count++;
}
}
}
}