Timus 1880. Psych Up's Eigenvalues题解

At one of the contests at the Petrozavodsk Training Camp, players of the Psych Up team quickly found the simple problem and Fedya sat at the computer. The solution was ready in five minutes. Without spending time for testing, Fedya submitted it and in a few seconds got Time Limit Exceeded.

Fedya crumpled the problem statement and left the room slamming the door behind him. Things were going the wrong way and he needed to wind down. On his way to the restroom Fedya heard a conversation of the contest authors. Pasha was discussing with his friend the solution of the problem for which Fedya had got TLE. Fedya could only discern the word “eigenvalues.”

Fedya thought about it and decided that he, for sure, had his eigenvalues. For example, the date of birth, the number of his apartment, the mark he had got at the latest exam, or the number of travels to contests. But they had a team contest, so what was their team's eigenvalue? Of course, a number was a team's eigenvalue if it was an eigenvalue for each of its players. With these joyful thoughts Fedya returned to the contest room.

Input

The input data consist of three blocks two lines each. The first line of each block contains the numbernof a player's eigenvalues<nobr>(1 ≤<em>n</em>≤ 4000)</nobr>. In the second line you are givenndistinct integers in ascending order, which are the eigenvalues. All the eigenvalues are positive integers not exceeding109.

Output

Output the number of eigenvalues of the Psych Up team.

Sample

input	output
5 13 20 22 43 146 4 13 22 43 146 5 13 43 67 89 146	3

即比较三个数组有多少相同的数字：

可以时间效率是O(n)，n是最小数组的长度

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

void PsychUpEigenvalues()
{
	int t1 = 0, t2 = 0, t3 = 0;
	cin>>t1;
	vector<int> v1(t1);
	for (int i = 0; i < t1; i++)
	{
		cin>>v1[i];
	}
	cin>>t2;
	vector<int> v2(t2);
	for (int i = 0; i < t2; i++)
	{
		cin>>v2[i];
	}
	cin>>t3;
	vector<int> v3(t3);
	for (int i = 0; i < t3; i++)
	{
		cin>>v3[i];
	}
	int c = 0;
	for (int i = 0, j = 0, k = 0; i < t1 && j < t2 && k < t3; )
	{
		int t = min(v1[i], min(v2[j], v3[k]));
		if (t == v1[i] && t == v2[j] && t == v3[k])
		{
			c++, i++, j++, k++;
		}
		else if (t == v1[i]) i++;
		else if (t == v2[j]) j++;
		else k++;
	}
	cout<<c<<endl;
}