思维题 HDOJ 5288 OO’s Sequence

题目传送门

 /*
     定义两个数组，l[i]和r[i]表示第i个数左侧右侧接近它且值是a[i]因子的位置，
         第i个数被选择后贡献的值是(r[i]-i)*(i-l[i])，每个数都枚举它的因子，更新l[i], r[i]，复杂度O(n*sqrt(a[i]))
 */
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <cstring>
 #include <map>
 using namespace std;

 const int MAXN = 1e5 + ;
 const int INF = 0x3f3f3f3f;
 const int MOD = 1e9 + ;
 int a[MAXN];
 int l[MAXN], r[MAXN];
 int pos[];

 int main(void)  {       //HDOJ 5288 OO’s Sequence
     //freopen ("1001.in", "r", stdin);
     int n;
     while (scanf ("%d", &n) == )   {
         for (int i=; i<=n; ++i)    {
             scanf ("%d", &a[i]);
         }

         int ans = ;
         memset (pos, , sizeof (pos));
         memset (l, , sizeof (l));
         for (int i=; i<=n; ++i)    {
             for (int j=; j<=sqrt (a[i]); ++j)    {
                 if (a[i] % j == )  {
                     if (pos[j] > l[i])  l[i] = pos[j];
                     if (pos[a[i]/j] > l[i])    l[i] = pos[a[i]/j];
                 }
             }
             pos[a[i]] = i;
         }
         for (int i=; i<=; ++i)    pos[i] = n + ;
         for (int i=; i<=n; ++i)    r[i] = n + ;
         for (int i=n; i>=; --i)    {
             for (int j=; j<=sqrt (a[i]); ++j)    {
                 if (a[i] % j == )  {
                     if (pos[j] < r[i])  r[i] = pos[j];
                     if (pos[a[i]/j] < r[i])    r[i] = pos[a[i]/j];
                 }
             }
             pos[a[i]] = i;
         }
         for (int i=; i<=n; ++i)    {
             ans = (ans + (r[i] - i) * (i - l[i]) % MOD) % MOD;
         }
         printf ("%d\n", ans);
     }

     return ;
 }