task push_back rip cas inline inpu color pac input

Discription 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.



預處理一下每個數的約數，直接暴力做就行了。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<vector>
#include<cstring>
#define ll long long
#define maxn 100005
#define pb push_back
using namespace std;
ll tot=0;
vector<int> son[maxn];
int n,m,a[maxn],f[maxn];
int mult[maxn],g[maxn],to;

inline void init(){
    for(int i=1;i<=100000;i++)
        for(int j=i;j<=100000;j+=i) son[j].pb(i);
}

int main(){
    init();
    
    while(scanf("%d",&n)==1&&n){
        memset(mult,0,sizeof(mult));
        for(int i=1;i<=n;i++){
            scanf("%d",a+i);
            f[i]=mult[a[i]];
            if(!f[i]) f[i]=i;
            for(int j=son[a[i]].size()-1;j>=0;j--){
                to=son[a[i]][j];
                mult[to]=max(mult[to],i);
            }
        }

        memset(mult,0x3f,sizeof(mult));
        for(int i=n;i;i--){
            g[i]=mult[a[i]];
            if(g[i]==mult[0]) g[i]=i;
            for(int j=son[a[i]].size()-1;j>=0;j--){
                to=son[a[i]][j];
                mult[to]=min(mult[to],i);
            }            
        }
        
        tot=0;
        for(int i=1;i<=n;i++) tot+=(ll)a[f[i]]*(ll)a[g[i]];
        printf("%lld\n",tot);
    }
    
    return 0;
}

HDOJ 4961 Boring Sum