pku 1855 Mint 求4个数的最小公倍数，欧几里得算法

pku 1855 Mint 求4个数的最小公倍数，欧几里得算法

题目很长，就是说用从n种硬币中选取4种不同的硬币堆叠起来，作为桌子的四个腿，显然，桌子的四个腿要相等。问对于一个给定的高度h，硬币能够堆成的不大于h的最大高度和不小于h的最小高度分别为多少？
应为数据不大， n为50左右，可以用4个for循环来枚举硬币，假设四种硬币为a、b、c、d，求的这四个数的最小公倍数即可，然后就用欧几里得算法解决。

 1 /**/ /*
 2Source Code
 3
 4Problem: 1855  User: yzhw 
 5Memory: 8648K  Time: 313MS 
 6Language: G++  Result: Accepted 
 7*/
 8
 9
10 # include  < iostream >
11 # include  < cstdio >
12 using   namespace  std;
13 int  l[ 51 ];
14 int  refer[ 51 ][ 51 ][ 51 ][ 51 ];
15 int  gcd( int  a, int  b)
16 {
17    while(b)
18    {
19        int t=a%b;
20        a=b;
21        b=t;
22    }
23    return a;
24}
25 int  main()
26 {
27    int n,t;
28    while(true)
29    {
30        scanf("%d%d",&n,&t);
31        if(!n&&!t) break;
32        for(int i=0;i<n;i++)
33            scanf("%d",l+i);
34        for(int a=0;a<n;a++)
35                                for(int b=a+1;b<n;b++)
36                                    //if(a!=b)
37                                    for(int c=b+1;c<n;c++)
38                                        //if(a!=c&&b!=c)
39                                        for(int d=c+1;d<n;d++)
40                                            //if(a!=d&&b!=d&&c!=d)
41                                            {
42                                                int lcd1=l[a]*l[b]/gcd(l[a],l[b]),lcd2=l[c]*l[d]/gcd(l[c],l[d]);
43                                                int lcd=lcd1*lcd2/gcd(lcd1,lcd2);
44                                                refer[a][b][c][d]=lcd;
45
46                                            }
47
48        while(t--)
49        {
50            int minnum=0xfffffff,maxnum=-1;
51            int pos;
52            scanf("%d",&pos);
53            for(int a=0;a<n;a++)
54                for(int b=a+1;b<n;b++)
55                    for(int c=b+1;c<n;c++)
56                        for(int d=c+1;d<n;d++)
57                            {
58
59                                int lcd=refer[a][b][c][d];
60                                if(pos-pos%lcd>maxnum)
61                                    maxnum=pos-pos%lcd;
62                                if(pos%lcd&&lcd*(pos/lcd+1)<minnum)
63                                    minnum=lcd*(pos/lcd+1);
64                                if(pos%lcd==0&&pos<minnum)
65                                    minnum=pos;
66
67                            }
68            printf("%d %d\n",maxnum,minnum);
69        }
70    }
71    return 0;
72}
73
74