Telephone madman algorithm problem ---

Telephone madman

Title Description

Given a large number of mobile phone users call records to find out where the most number of calls chat madman.

Entry

Firstly, input a positive integer N (≤105), the number of records for the call. Then N lines of a given call record. For simplicity, just to name a dialed phone number and the recipient's 11 digit number, wherein separated by spaces.

4
13005711862 13588625832
13505711862 13088625832
13588625832 18087925832
15005713862 13588625832

Export

Maniac given chat phone number in a row and the number of calls, separated by a space therebetween. If such a person is not unique, the minimum number and the number of calls in a madman output, and given the additional number of parallel madman.

13588625832 3

Thinking

Telephone number is very long, with the long long storage type, a first row of the array sequence, then find their number

Code

#include<iostream>
#include<algorithm>
using namespace std;  

int main()
{
 	long long a[220],b[220];//不带要删除了，就整了两个数组
 	int n,j=0,num[220]={1},max=0;
 	cin>>n;
 	for(int i=0;i<2*n;i++)
 	{
 		cin>>a[i];
 	}
 	fill(num,num+2*n,1); //现将计数数组全部赋值为1
 	sort(a,a+2*n); //排序，这样更好找一样的号码
 	
 	for(int i=0;i<2*n;i++) //计数
 	{
 		b[j]=a[i];
 		if(a[i+1]==a[i])
 		{
 			num[j]++;
 		}
 		else 
 		{	
		 	j++;
 		}
		 
 	}
 	for(int i=0;i<2*n;i++) //求最大的数
 	{
 		if(num[i]>max)
 		{
 			max=num[i];
 		}
 	}
 	for(int i=0;i<2*n;i++) //以防出现多个数量一样的号码
 	{
 		if(num[i]==max)
 		{
 			cout<<b[i]<<" "<<num[i]<<endl;
 		}
 	}
	return 0;
}