http://poj.org/problem?id=4044

大致题意：给出两个班级的成绩，先按降序排序，并且没有成绩相同的。然后求连续的最长公共子序列。输出时，先输出最长公共子序列，然后按个位数字递增的顺序输出，若各位数字一样就按成绩递增。

人数小于30，注意去重，直接暴力即可。

#include <stdio.h>
#include <iostream>
#include <map>
#include <stack>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#include <stdlib.h>
#include <algorithm>
#define LL long long
#define _LL __int64
#define eps 1e-8
#define PI acos(-1.0)
using namespace std;

const int maxn = 32;
int n1,n2;
int a[maxn],b[maxn],aa[maxn],bb[maxn];

int cmp(int a, int b)
{
	return a > b;
}

struct node
{
	int num;
	int dig;

	bool operator < (const struct node &tmp)const
	{
		if(dig == tmp.dig)
			return num < tmp.num;
		return dig < tmp.dig;
	}
}ans[maxn];

bool judge(int s1, int s2, int len)
{
	int k;
	for(k = 0; k < len; k++)
	{
		if(a[s1+k] != b[s2+k])
			break;
	}
	if(k < len)
		return false;
	return true;
}

int main()
{
	int test;
	scanf("%d",&test);
	while(test--)
	{
		int i,j,t;
		scanf("%d %d",&n1,&n2);
		for(i = 0; i < n1; i++)
			scanf("%d",&aa[i]);
		for(i = 0; i < n2; i++)
			scanf("%d",&bb[i]);

		sort(aa,aa+n1,cmp);
		sort(bb,bb+n2,cmp);

		a[0] = aa[0];
		t = 1;
		for(i = 1; i < n1;)
		{
			while(aa[i] == aa[i-1] && (i+1) < n1)
				i++;
			if(aa[i] != aa[i-1])
				a[t++] = aa[i++];
			else break;
		}
		n1 = t;

		b[0] = bb[0];
		t = 1;
		for(i = 1; i < n2; )
		{
			while(bb[i] == bb[i-1] && (i+1) < n2)
				i++;
			if(bb[i] != bb[i-1])
				b[t++] = bb[i++];
			else break;
		}
		n2 = t;
		
		int len = 0;
		int cnt;

		for(i = 0; i < n1; i++)
		{
			for(j = 0; j < n2; j++)
			{
				for(int k = 1; k <= min(n1-i,n2-j); k++)
				{
					if(judge(i,j,k) && len < k)
					{
						len = k;
						cnt = 0;
						for(int g = 0; g < k; g++)
							ans[cnt++] = (struct node){a[i+g],a[i+g]%10};
					}
				}
			}
		}

		if(len == 0)
		{
			printf("NONE\n");
			continue;
		}

		for(int i = 0; i < cnt-1; i++)
			printf("%d ",ans[i].num);
		printf("%d\n",ans[cnt-1].num);

		sort(ans,ans+cnt);
		for(i = 0; i < cnt-1; i++)
			printf("%d ",ans[i].num);
		printf("%d\n",ans[cnt-1].num);

	}
	return 0;
}