HDOJ 5288 OO’s Sequence 水

预处理出每个数字的左右两边可以整除它的最近的数的位置

OO’s Sequence

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

Total Submission(s): 1880    Accepted Submission(s): 672


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

 

Author

FZUACM

 

Source

2015 Multi-University Training Contest 1

 

/* ***********************************************
Author        :CKboss
Created Time  :2015年07月24日 星期五 08时12分15秒
File Name     :HDOJ5288.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

typedef long long int LL;

const int maxn=100100;
const LL MOD=(LL)(1e9+7);

int n;
int a[maxn];
int Left[maxn],Right[maxn];
vector<int> pos[10010];

void init()
{
for(int i=0;i<=10010;i++)
{
pos[i].clear();
}
for(int i=0;i<n;i++)
{
Left[i]=0; Right[i]=n-1;
}
}

void pre()
{
for(int i=0;i<n;i++)
{
int x=a[i];
for(int j=x;j<=10000;j+=x)
{
for(int k=0,sz=pos[j].size();k<sz;k++)
{
int z=pos[j][k];
if(z==i) continue;
else if(z<i)
{
Right[z]=min(Right[z],i-1);
}
else if(z>i)
{
Left[z]=max(Left[z],i+1);
}
}
}
}
}

int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);

while(scanf("%d",&n)!=EOF)
{
init();
for(int i=0;i<n;i++)
{
scanf("%d",a+i);
pos[a[i]].push_back(i);
}
pre();
LL ans=0;
for(int i=0;i<n;i++)
{
LL L=i-Left[i]+1LL;
LL R=Right[i]-i+1LL;
ans=(ans+(L*R)%MOD)%MOD;
}
cout<<ans<<endl;
}

return 0;
}