题目链接：

Problem Description

Number theory is interesting, while this problem is boring.

Here is the problem. Given an integer sequence a

₁, a

₂, …, a

_n, let S(i) = {j|1<=j<i, and a

_j is a multiple of a

_i}. If S(i) is not empty, let f(i) be the maximum integer in S(i); otherwise, f(i) = i. Now we define bi as a

_f(i). Similarly, let T(i) = {j|i<j<=n, and a

_j is a multiple of a

_i}. If T(i) is not empty, let g(i) be the minimum integer in T(i); otherwise, g(i) = i. Now we define c

_i as a

_g(i). The boring sum of this sequence is defined as b

₁ * c

₁ + b

₂ * c

₂ + … + b

_n * c

_n.

Given an integer sequence, your task is to calculate its boring sum.

 

Input

The input contains multiple test cases.

Each case consists of two lines. The first line contains an integer n (1<=n<=100000). The second line contains n integers a

₁, a

₂, …, a

_n (1<= a

_i<=100000).

The input is terminated by n = 0.

 

Output

Output the answer in a line.

 

Sample Input

5 1 4 2 3 9 0

 

Sample Output

136

Hint

In the sample, b1=1, c1=4, b2=4, c2=4, b3=4, c3=2, b4=3, c4=9, b5=9, c5=9, so b1 * c1 + b2 * c2 + … + b5 * c5 = 136.

 

Author

SYSU

 

Source

题意：

给出一个数列：a[i]，然后

b[i]：表示在 i 前面的项，假设有a[i]的倍数（要最靠近i的），那么b[i]就等于这个数，假设没有那么b[i] = a[i];

c[i]：表示在 i 后面的项，假设有a[i]的倍数（要最靠近i的），那么c[i] 就等于这个数，假设没有那么c[i] = a[i];

思路：

//先打表，把每一个数的约数存在vector里；

//然后从前往后扫一遍，结果存在b[i]，

//Ps：假设不清楚为什么从前往后扫一遍就是最靠近的那个数可调试一下（案例：9 6 3 2 1）；

//然后从后往前扫一遍，结果存在c[i]，

//Ps：假设不清楚为什么从后往前扫一遍就是最靠近的那个数可调试一下（案例：1 2 3 6 9）；

//最后计算b[i]*c[i]的和就可以。

代码例如以下：

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #define maxn 100000+17using namespace std;typedef __int64 LL;int vis[maxn];int a[maxn], b[maxn], c[maxn];vector
         
          V[maxn];void init(){    for(int i = 0; i < maxn; i++)        V[i].clear();    for(int i = 1; i <= maxn; i++)    {        for(int j = 1; j*i <= maxn; j++)//每一个数对应的约数        {            V[i*j].push_back(i);//i是哪些数的约数        }    }}int main(){    int n;    init();    while(scanf("%d",&n) && n)    {        for(int i = 1; i <= n; i++)        {            scanf("%d",&a[i]);        }        memset(vis,0,sizeof(vis));        for(int i = 1; i <= n; i++)        {            if(vis[a[i]] == 0)                b[i]=a[i];            else                b[i]=vis[a[i]];//a[i]的倍数            for(int j = 0; j < V[a[i]].size(); j++)                vis[V[a[i]][j]] = a[i];//V[a[i]][j]为a[i]的约数        }        memset(vis,0,sizeof(vis));        for(int i = n; i >= 1; i--)        {            if(vis[a[i]] == 0)                c[i] = a[i];            else                c[i] = vis[a[i]];            for(int j = 0; j < V[a[i]].size(); j++)                vis[V[a[i]][j]] = a[i];        }        LL sum=0;        for(int i = 1; i <= n; i++)        {            sum += (LL)b[i]*(LL)c[i];        }        printf("%I64d\n",sum);    }    return 0;}