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 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[MAXN];
int l[MAXN], r[MAXN];
int pos[10010];

int main(void)  {       //HDOJ 5288 OO’s Sequence
    //freopen ("1001.in", "r", stdin);
    int n;
    while (scanf ("%d", &n) == 1)   {
        for (int i=1; i<=n; ++i)    {
            scanf ("%d", &a[i]);
        }
        
        int ans = 0;
        memset (pos, 0, sizeof (pos));
        memset (l, 0, sizeof (l));
        for (int i=1; i<=n; ++i)    {
            for (int j=1; j<=sqrt (a[i]); ++j)    {
                if (a[i] % j == 0)  {
                    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=1; i<=10000; ++i)    pos[i] = n + 1;
        for (int i=1; i<=n; ++i)    r[i] = n + 1;
        for (int i=n; i>=1; --i)    {
            for (int j=1; j<=sqrt (a[i]); ++j)    {
                if (a[i] % j == 0)  {
                    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=1; i<=n; ++i)    {
            ans = (ans + (r[i] - i) * (i - l[i]) % MOD) % MOD;
        }
        printf ("%d
", ans);
    }

    return 0;
}