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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


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 a_i mod a_j=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 a_i(0<a_i<=10000)

Output For each tests: ouput a line contain a number ans.
Sample Input

5
1 2 3 4 5

Sample Output

23

Author FZUACM
Source 2015 Multi-University Training Contest 1
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#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;
}

HDU 5288(OO’s Sequence-區間互質情況統計)