暴力+降复杂度 BestCoder Round #39 1002 Mutiple

题目传送门

 /*
     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，
         若没有，默认加0，nlogn复杂度；
     我用暴力竟然水过去了:)
 */
 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <string>
 #include <algorithm>
 using namespace std;

 const int MAXN = 1e4 + ;
 const int INF = 0x3f3f3f3f;
 int a[MAXN], b[MAXN];

 int main(void)        //BestCoder Round #39 1002 Mutiple
 {
     //freopen ("1002.in", "r", stdin);

     int n;

     while (scanf ("%d", &n) == )
     {
         memset (b, , sizeof (b));
         for (int i=; i<=n; ++i)    scanf ("%d", &a[i]);

         long long ans = ;
         for (int i=n; i>=; --i)
         {
             ans += b[a[i]];
             for (int j=; j*j<=a[i]; ++j)
             {
                 if (a[i] % j == )
                 {
                     b[j] = i;    b[a[i]/j] = i;
                 }
             }
         }

         printf ("%lld\n", ans);
     }

     return ;
 }

 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <string>
 #include <algorithm>
 using namespace std;

 const int MAXN = 1e4 + ;
 const int INF = 0x3f3f3f3f;
 int a[MAXN];

 int main(void)        //BestCoder Round #39 1002 Mutiple
 {
     //freopen ("1002.in", "r", stdin);

     int n;
     long long sum = ;

     while (scanf ("%d", &n) == )
     {
         for (int i=; i<=n; ++i)    scanf ("%d", &a[i]);

         sum = ;    int k = ;
         for (int i=; i<=n-; ++i)
         {
             k = ;
             for (int j=i+; j<=n; ++j)
             {
                 if (a[j] % a[i] == )
                 {
                     k = j;    break;
                 }
             }
             sum += k;
         }

         printf ("%lld\n", sum);
     }

     return ;
 }

brute