hdu 5288 OO’s Sequence

题目链接：​​http://acm.hdu.edu.cn/showproblem.php?pid=5288​​

解法：
定义两个数组L[i],R[i],表示第i数左侧和右侧最接近它且值是a[i]因子的数字的位置，那么第i个数能贡献的答案就是(R[i]-i)*(i-L[i]),因此每个数字x都去枚举它的因子y，然后左右找到一个值是y且最接近x的数，然后用他的位置更新一下L,R数组。时间复杂度O(nsqrt(a))。

代码：

#include <stdio.h>
#include <ctime>
#include <math.h>
#include <limits.h>
#include <complex>
#include <string>
#include <functional>
#include <iterator>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <bitset>
#include <sstream>
#include <iomanip>
#include <fstream>
#include <iostream>
#include <ctime>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <time.h>
#include <ctype.h>
#include <string.h>
#include <string>
#include <assert.h>
#pragma comment(linker,"/STACK:1024000000,1024000000")

using namespace std;

const int MAXN = 100010;
const int MOD = 1e9 + 7;

int n;
int a[MAXN];
int L[MAXN], R[MAXN];
int pre[MAXN];
vector<int> g[MAXN];

void init()
{
    for (int i = 1; i <= 10000; i++)
    {
        g[i].clear();
        for (int j = 1; j*j <= i; j++)
        {
            if (i%j == 0)
            {
                g[i].push_back(j);
                if (j*j != i)
                    g[i].push_back(i / j);
            }
        }
    }
}

int main()
{
    init();
    while (~scanf("%d", &n))
    {
        for (int i = 1; i <= n; i++)
            scanf("%d",&a[i]);

        memset(L,0,sizeof(L));
        memset(R,0,sizeof(R));
        memset(pre,0,sizeof(pre));

        for (int i = 1; i <= n; i++)
        {
            int tmp = a[i];
            int pos = 0;
            for (int j = 0; j < g[tmp].size(); j++)
            {
                int t2 = g[tmp][j];
                pos = max(pre[t2],pos);
            }
            pre[tmp] = i;
            L[i] = pos;
        }
        for (int i = 1; i <= 10000; i++) pre[i] = 100001;

        for (int i = n; i >= 1;i--)
        {
            int tmp = a[i];
            int pos = n + 1;
            for (int j = 0; j < g[tmp].size(); j++)
            {
                int t2 = g[tmp][j];
                pos = min(pre[t2], pos);
            }
            pre[tmp] = i;
            R[i] = pos;
        }
        long long ans = 0;
        for (int i = 1; i <= n; i++)
            ans += (long long)(R[i] - i) * (i - L[i]) % MOD;
        printf("%lld\n", ans%MOD);
    }
    return 0;
}