2015多校联赛 ——HDU5288（数学）

OO’s Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 449    Accepted Submission(s): 158

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 （10^9+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

 

Source

2015 Multi-University Training Contest 1

求在[a , b] 中有多少个i 满足找不到另一个数j 使,a[i] % a[j] = 0

用L[]，R[]分别记录左右离起最近的因子，ans+=(R[i]-i+1)*(i-L[i]+1)/2 %MOD

//自己并没有想到怎么做 (╯▽╰)

#include<cstdio>#include<cstring>#include<iostream>using namespace std ;const int maxn = 100010 ;const int mod = 1e9+7 ;int a[maxn];long long  last[maxn] ;long long pre[maxn] ;long long  l[maxn] ;long long r[maxn] ;int  main(){    //freopen("1001.txt" ,"r" ,stdin) ;    int n;    while(~scanf("%d",&n))    {        memset(pre,0,sizeof(pre));        memset(last,0,sizeof(last));        for(int i = 1;i <= n;i++)        {            l[i] = 1;            r[i] = n;            scanf("%d",&a[i]);            for(int j = a[i];j < 10001;j+=a[i])            {                if(pre[j] && r[pre[j]] == n)                    r[pre[j]] = i-1;                pre[a[i]] = i;            }        }        for(int i = n;i >=1;i--)        {            for(int j = a[i];j < 10001;j+=a[i])            {                if(last[j] && l[last[j]]==1)                    l[last[j]] = i+1;                last[a[i]] = i;            }        }        long long ans = 0;        for(long long int i = 1;i <= n;i++)            ans= (ans+(((i-l[i]+1)%mod)*((r[i] - i +1)%mod))%mod)%mod;        printf("%I64d\n",ans);    }    return 0 ;}