暴力+降复杂度 BestCoder Round #39 1002 Mutiple

题目传送门

 1 /*
 2     设一个b[]来保存每一个a[]的质因数的id，从后往前每一次更新质因数的id，
 3         若没有，默认加0，nlogn复杂度；
 4     我用暴力竟然水过去了:)
 5 */
 6 #include <cstdio>
 7 #include <iostream>
 8 #include <cstring>
 9 #include <string>
10 #include <algorithm>
11 using namespace std;
12
13 const int MAXN = 1e4 + 10;
14 const int INF = 0x3f3f3f3f;
15 int a[MAXN], b[MAXN];
16
17 int main(void)        //BestCoder Round #39 1002 Mutiple
18 {
19     //freopen ("1002.in", "r", stdin);
20
21     int n;
22
23     while (scanf ("%d", &n) == 1)
24     {
25         memset (b, 0, sizeof (b));
26         for (int i=1; i<=n; ++i)    scanf ("%d", &a[i]);
27
28         long long ans = 0;
29         for (int i=n; i>=1; --i)
30         {
31             ans += b[a[i]];
32             for (int j=1; j*j<=a[i]; ++j)
33             {
34                 if (a[i] % j == 0)
35                 {
36                     b[j] = i;    b[a[i]/j] = i;
37                 }
38             }
39         }
40
41         printf ("%lld\n", ans);
42     }
43
44     return 0;
45 }

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <string>
 5 #include <algorithm>
 6 using namespace std;
 7
 8 const int MAXN = 1e4 + 10;
 9 const int INF = 0x3f3f3f3f;
10 int a[MAXN];
11
12 int main(void)        //BestCoder Round #39 1002 Mutiple
13 {
14     //freopen ("1002.in", "r", stdin);
15
16     int n;
17     long long sum = 0;
18
19     while (scanf ("%d", &n) == 1)
20     {
21         for (int i=1; i<=n; ++i)    scanf ("%d", &a[i]);
22
23         sum = 0;    int k = 0;
24         for (int i=1; i<=n-1; ++i)
25         {
26             k = 0;
27             for (int j=i+1; j<=n; ++j)
28             {
29                 if (a[j] % a[i] == 0)
30                 {
31                     k = j;    break;
32                 }
33             }
34             sum += k;
35         }
36
37         printf ("%lld\n", sum);
38     }
39
40     return 0;
41 }

brute