HDU 5288 OO’s Sequence（数学啊 多校2015）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5288

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

51 2 3 4 5

 

Sample Output

23

 

Author

FZUACM

 

Source

2015 Multi-University Training Contest 1

题意：

给出n个数，让找到所有的区间内不能整除其它数的数的个数之和 ！

PS:

代码如下：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define maxn 100017#define LL __int64const int mod = 1e9+7;LL l[maxn], r[maxn];//存储左边因子，右边因子的位置LL a[maxn];int main(){    int n;    LL pre[maxn], last[maxn];    while(~scanf("%d",&n))    {        for(int i = 1; i <= n; i++)        {            scanf("%I64d",&a[i]);            l[i] = 1;            r[i] = n;//初始化最左边的因子和最右边的因子都是本身        }        memset(pre,0,sizeof(pre));        memset(last,0,sizeof(last));        for(int i = 1; i <= n; i++)        {            for(int j = a[i]; j <= 10000; j+=a[i])//枚举a[i]的倍数            {                if(pre[j]!=0 && r[pre[j]] == n)//如果j已经出现并且在右边最近的因子还没有找到                {                    r[pre[j]] = i-1;                }            }            pre[a[i]] = i;        }        for(int i = n; i >= 1; i--)        {            for(int j = a[i]; j <= 10000; j+=a[i])//枚举a[i]的倍数            {                if(last[j]!=0 && l[last[j]] == 1)//如果j已经出现并且在左边最近的因子还没有找到                {                    l[last[j]] = i+1;                }            }            last[a[i]] = i;        }//        for(int i=1;i<=n;i++){//            printf("%d %I64d %I64d  %I64d\n",i,l[i],r[i],(i-l[i]+1)*(r[i]-i+1));//        }        LL ans = 0;        for(int i = 1; i <= n; i++)        {            ans+=((LL)(i-l[i]+1)*(r[i]-i+1))%mod;            ans%=mod;        }        printf("%I64d\n",ans);    }    return 0;}