TLE：csu 1016 最小公倍数

求删改方案总共多少种，预料之中的超时，因为所有提交的22次中有12次超时。

 1 # include <stdio.h>
 2 
 3 # define MAXN 205
 4 
 5 int gcd(int a, int b);
 6 
 7 int m[MAXN];
 8 int f[MAXN][MAXN];
 9 
10 int main()
11 {
12      int i, j, T, n, lcm, tmp, cnt;
13      
14      scanf("%d", &T);
15      while (T--)
16      {
17            cnt = 0;
18            lcm = 1;
19            scanf("%d", &n);
20            for (i = 1; i <= n; ++i)
21            {                   
22                    scanf("%d", &m[i]);
23                    lcm /= gcd(lcm, m[i]);
24                    lcm *= m[i];
25              }
26              for (i = 1; i < n; ++i)
27              for (j = i+1; j <= n; ++j)
28                    f[i][j] = f[j][i] = gcd(m[i], m[j]);
29            for (i = 1; i <= n; ++i)
30            {
31                  for (tmp = m[i], j = 1; j <= n; ++j)
32                        if(i != j && 1 == f[i][j])
33                           tmp *= m[j];
34                  if (tmp == lcm) 
35                  {
36                           for (j = 1; j <= n; ++j)
37                                if (i != j && 1 == f[i][j])
38                                    f[j][i] = f[i][j] = -1;
39                           ++cnt;
40                     }
41             }
42             printf("%d\n",  cnt);
43       }
44      
45      return 0;
46 }
47 
48 int gcd(int x, int y)
49 {
50      while (x != y)
51      {
52            if (x < y) y = y - x;
53            else x = x - y;
54       }
55       return x;
56 }

// 待续