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

5
1 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 __int64
const 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;
}

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