OO’s Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 2643    Accepted Submission(s): 925

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 a_i mod a_j=0,now OO want to know

∑i=1n∑j=inf(i,j) mod （109+7）.

5

1 2 3 4 5

23

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<algorithm>

#include<functional>

#include<iostream>

#include<cmath>

#include<cctype>

#include<ctime>

#include<vector>

using namespace std;

#define For(i,n) for(int i=1;i<=n;i++)

#define Fork(i,k,n) for(int i=k;i<=n;i++)

#define Rep(i,n) for(int i=0;i<n;i++)

#define ForD(i,n) for(int i=n;i;i--)

#define RepD(i,n) for(int i=n;i>=0;i--)

#define Forp(x) for(int p=pre[x];p;p=next[p])

#define Forpiter(x) for(int &p=iter[x];p;p=next[p])

#define Lson (x<<1)

#define Rson ((x<<1)+1)

#define MEM(a) memset(a,0,sizeof(a));

#define MEMI(a) memset(a,127,sizeof(a));

#define MEMi(a) memset(a,128,sizeof(a));

#define INF (2139062143)

#define F (1000000007)

#define MAXN (1000000+10)

#define MAXn (1000000+10)

typedef long long ll;

ll mul(ll a,ll b){return (a*b)%F;}

ll add(ll a,ll b){return (a+b)%F;}

ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}

void upd(ll &a,ll b){a=(a%F+b%F)%F;}

int a[MAXn],n;

ll l[MAXN],r[MAXN];

ll al[MAXN],ar[MAXN];

int main()

{

//	freopen("A.in","r",stdin);

//	freopen(".out","w",stdout);

	while(scanf("%d",&n)==1)

	{

		ll ans=0;

		For(i,n) scanf("%d",&a[i]);

		MEM(l) MEMI(r)

		For(i,n)

		{

			al[i]=0;

			int p=a[i];

			for(int j=1;(ll)j*j<=(ll)p;j++) {

				if (p%j==0) al[i]=max(al[i],max(l[j],l[p/j]));

			}

			l[a[i]]=i;

		}

		ForD(i,n)

		{

			ar[i]=n+1;

			int p=a[i];

			for(int j=1;(ll)j*j<=(ll)p;j++) {

				if (p%j==0) ar[i]=min(ar[i],min(r[j],r[p/j]));

			}

			r[a[i]]=i;

		}

// 		For(i,n) cout<<al[i]<<' ';cout<<endl;

// 		For(i,n) cout<<ar[i]<<' ';cout<<endl;

//		

		For(i,n)

			upd(ans,mul(i-al[i],ar[i]-i));

		cout<<ans<<endl;

	}	

	return 0;

}