OO’s Sequence

题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=5288

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

Hint

题意

f(l,r)表示[l,r]区间中有多少个i满足在这个区间中找不到其他j使得ai%aj=0

然后让你输出所有f(l,r)的累加。

题解：

对于每一个位置，我算贡献就好了。

直接暴力分解a[i]就好了。

复杂度nsqrtn的。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+6;
const int mod = 1e9+7;
int a[maxn],n;
int L[maxn],R[maxn];
int p[maxn];
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        memset(L,0,sizeof(L));
        memset(R,0,sizeof(R));
        memset(p,0,sizeof(p));
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=sqrt(a[i]);j++)
                if(a[i]%j==0)L[i]=max(L[i],p[j]+1),L[i]=max(L[i],p[a[i]/j]+1);
            p[a[i]]=i;
        }
        reverse(a+1,a+1+n);
        memset(p,0,sizeof(p));
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=sqrt(a[i]);j++)
                if(a[i]%j==0)R[i]=max(R[i],p[j]+1),R[i]=max(R[i],p[a[i]/j]+1);
            p[a[i]]=i;
        }
        long long ans = 0;
        for(int i=1;i<=n;i++)
        {
            ans += 1ll*(i-L[i]+1)*(n-R[n-i+1]+1-i+1);
            ans%=mod;
            //cout<<L[i]<<" "<<n-R[i]+1<<endl;
        }
        cout<<ans<<endl;
    }
}