HDU 5288——OO’s Sequence——————【技巧题】

OO’s Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1549    Accepted Submission(s): 559

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

 

 

题目大意：给你一个长度为n的区间。求这个区间内任意子区间中ai没有因子的ai的个数。

 

解题思路：从左往右遍历a数组得到L数组。对于ai，枚举j  1-->sqrt（ai），如果j是ai的约数并且j已经在ai之前，那么我用max取离i最近的位置；相应的ai/j也是ai的约数，同理如果ai/j也在ai之前，也用max取离i最近的位置。对于R数组，同理可得。

 

#include<stdio.h>

#include<string.h>

#include<math.h>

#include<algorithm>

using namespace std;

#define min(a,b) ((a)<(b)?(a):(b))

#define max(a,b) ((a)>(b)?(a):(b))

const int maxn=1e5+20;

const int INF=0x3f3f3f3f;

const int MOD=1e9+7;

int a[maxn],L[maxn],R[maxn];

int pos[maxn];

int main(){

    int n,limj;

    while(scanf("%d",&n)!=EOF){

        for(int i=1;i<=n;i++){

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

        }

        memset(pos,0,sizeof(pos));

        for(int i=1;i<=n;i++){

            L[i]=0;

            limj=(int)sqrt(a[i]);

            for(int j=1;j<=limj;j++){

                if(a[i]%j==0){

                    if(pos[j]){

                        L[i]=max(L[i],pos[j]);

                    }

                    if(pos[a[i]/j]){

                        L[i]=max(L[i],pos[a[i]/j]);

                    }

                }

            }

            pos[a[i]]=i;

        }

        memset(pos,0,sizeof(pos));

        for(int i=n;i>=1;i--){

            R[i]=n+1;

            limj=(int)sqrt(a[i]);

            for(int j=1;j<=limj;j++){

                if(a[i]%j==0){

                    if(pos[j]){

                        R[i]=min(R[i],pos[j]);

                    }

                    if(pos[a[i]/j]){

                        R[i]=min(R[i],pos[a[i]/j]);

                    }

                }

            }

            pos[a[i]]=i;

        }

        int sum=0,res;

        for(int i=1;i<=n;i++){

            sum=(sum%MOD+((i-L[i])*(R[i]-i))%MOD)%MOD;

        }

        printf("%d\n",sum);

    }

    return 0;

}