hdu 5288 - OO’s Sequence 数学

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 ai mod aj=0,now OO want to know

∑i=1n∑j=inf(i,j) mod （109+710^9+7）.

Input

There are multiple test cases. Please process till EOF.

In each test case:

First line: an integer n(n<=105n<=10^5) indicating the size of array

Second line:contain n numbers ai(0<ai≤100000)

Output

For each tests: ouput a line contain a number ans.

Sample Input

5

1 2 3 4 5

Sample Output

23

思路：对于每个数，考虑他能出现在哪些区间里，假如左区间长度为l，右区间长度为r，那么它的数量就是l*r，所以求出每个数的左右区间即可，可以通过枚举他的因子的位置获得

[code]#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn=100010;
const int INF=1000000010;
const int MOD=1e9+7;
int N,a[maxn];
int vis[maxn];
int L[maxn],R[maxn];
int main()
{
    while(scanf("%d",&N)!=EOF)
    {
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=N;i++)
            scanf("%d",&a[i]);
        for(int i=1;i<=N;i++)
        {
            int k=sqrt(a[i]);
            int tmp=-1;
            for(int j=1;j<=k;j++)
            {
                if(a[i]%j)continue;
                if(vis[j])tmp=max(tmp,vis[j]);
                if(j*j!=a[i]&&vis[a[i]/j])tmp=max(tmp,vis[a[i]/j]);
            }
            if(tmp==-1)L[i]=i;
            else L[i]=i-tmp;
            vis[a[i]]=i;
        }
        memset(vis,0,sizeof(vis));
        for(int i=N;i>=1;i--)
        {
            int k=sqrt(a[i]);
            int tmp=INF;
            for(int j=1;j<=k;j++)
            {
                if(a[i]%j)continue;
                if(vis[j])tmp=min(tmp,vis[j]);
                if(j*j!=a[i]&&vis[a[i]/j])tmp=min(tmp,vis[a[i]/j]);
            }
            if(tmp==INF)R[i]=N-i+1;
            else R[i]=tmp-i;
            vis[a[i]]=i;
        }
        LL ans=0;
        for(int i=1;i<=N;i++)
            (ans+=L[i]*R[i])%=MOD;
        printf("%I64d\n",ans);
    }
    return 0;
}