OO’s Sequence

Problem Description

OO has got a array A of size n ,defined a function f(l,r) represent the number of i (l<=i<=r) , that there's no j(l<=j<=r,j<>i) satisfy ai mod aj=0,now OO want to know  ∑i=1n∑j=inf(i,j) mod （109+7）.

Input

There are multiple test cases. Please process till EOF.
   In each test case: 
   First line: an integer n(n<=10^5)  indicating the size of array
   Second line:contain n numbers ai(0<ai<=10000)

Output

For each tests: ouput a line contain a number ans.

Sample Input

5
1 2 3 4 5

Sample Output

23

题意：

求每一个[L, R]区间中 j!=i a[i]对任意a[j]取余为0 的i的个数的总和

方法：

对于ai，看ai左边最近的位置l，al是ai的因数，右边位置r，ar是ai的因数。那么满足的个数就是（i-l)*(r-i)
预处理所有1-10000的因数。

从左到右，用pre[i]表示数字i出现的i的最右边的位置。算出每个位置的l。

同理从右到左，算出每个位置的r。

代码题解：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define maxn 100007
vector<int> head[10007];
int pre[10007];
int num[maxn];

int lef[maxn];
int righ[maxn];

int main()
{
    int n;
    for(int i = 1; i < 10000; i++)
    {
        head[i].clear();
        for(int j = 1; j * j <= i; j++)
        {
            if(i % j == 0)
            {
                head[i].push_back(j);
                head[i].push_back(i / j);          head[i]存储i的所有因子
            }
        }
    }
    while(scanf("%d", &n) != EOF)
    {
        for(int i = 1; i <= n; i++)
            scanf("%d", &num[i]);

        memset(pre, 0, sizeof(pre));
        for(int i = 1; i <= n; i++)
        {
            int u = num[i];
            int p = 0;
            for(int j = 0; j < head[u].size(); j++)
                p = max(p, pre[head[u][j]]);
            lef[i] = p;
            pre[u] = i;
        }

        memset(pre, 0x3f, sizeof(pre));
        for(int i = n; i > 0; i--)
        {
            int u = num[i];
            int p = n + 1;
            for(int j = 0; j < head[u].size(); j++)
                p = min(p, pre[head[u][j]]);
            righ[i] = p;
            pre[u] = i;
        }
        long long ans = 0, l, r;
        long long mod = 1000000007;
        for(int i = 1; i <= n; i++)
        {
            ans += (long long)(i - lef[i]) * (righ[i] - i);
            ans %= mod;
        }
        cout << ans << endl;
    }
    return 0;
}