HDU4961：Boring Sum

Problem Description

Number theory is interesting, while this problem is boring.

Here is the problem. Given an integer sequence a₁, a₂, …, 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 b&#31;i 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 &#31;c_i as a_g(i). The boring
sum of this sequence is defined as b₁ * c₁ + b₂ * c₂ + … + 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 a₁, a₂, …, 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.

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
#define up(i,x,y) for(i=x;i<=y;i++)
#define down(i,x,y) for(i=x;i>=y;i--)
#define mem(a,b) memset(a,b,sizeof(a))
#define w(x) while(x)
#define l 100005
#define ll __int64
#define s(a) scanf("%I64d",&a)
ll vis[l],a[l],b[l],c[l],n,i,j,ans;

int main()
{
    w((s(n),n))
    {
        up(i,1,n)
        s(a[i]);
        mem(vis,0);
        up(i,1,n)
        {
            if(vis[a[i]])
                b[i]=a[vis[a[i]]];
            else
                b[i]=a[i];
            up(j,1,sqrt(a[i]*1.0)+1)
            {
                if(a[i]%j==0)
                    vis[j]=vis[a[i]/j]=i;
            }
        }
        mem(vis,0);
        down(i,n,1)
        {
            if(vis[a[i]])
                c[i]=a[vis[a[i]]];
            else
                c[i]=a[i];
            up(j,1,sqrt(a[i]*1.0)+1)
            {
                if(a[i]%j==0)
                    vis[j]=vis[a[i]/j]=i;
            }
        }
        ans=0;
        up(i,1,n)
        ans+=b[i]*c[i];
        printf("%I64d\n",ans);
    }

    return 0;
}

HDU4961：Boring Sum