hdu 5288 （小技巧）

OO’s Sequence

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

 

題目大意： 給出n個數，求在任意區間內，i<>j 並且i%j！=0 的i的個數 

思        路： 利用pre數組存儲每個數出現的序號

                    借用此從左到右，便利出每個數據的因子，找到在數據範圍最右面的那個因子所在的序號

                   同理可知道左面的。

                  

 具體看代碼：

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int mod = 1e9+7;
#define ll long long
#define N 100005
ll left[N],right[N];
int n,num[N],pre[N];

int main()
{
    while(scanf("%d",&n)!=EOF){
        for(int i=1;i<=n;i++){
            scanf("%d",&num[i]);
            left[i]=1;
            right[i]=n;
        }
          
        memset(pre,0,sizeof(pre));
        for(int i=1;i<=n;i++){
            for(int j=num[i];j<=10000;j+=num[i]){
                if(pre[j]&&right[pre[j]]==n)
                    right[pre[j]]=i-1;
            }
            pre[num[i]]=i;
        }
        memset(pre,0,sizeof(pre));
        for(int i=n;i>=1;i--){
            for(int j=num[i];j<=10000;j+=num[i]){
                if(pre[j]&&left[pre[j]]==1)
                    left[pre[j]]=i+1;
            }
            pre[num[i]]=i;
        }

        long long ans=0;
        for(int i=1;i<=n;i++)
            ans = (ans%mod+(long long)(i-left[i]+1)*(right[i]-i+1)%mod)%mod;
        printf("%lld\n",ans);
    }
    return 0;
}