hdu 4961 Boring Sum

Boring Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 698    Accepted Submission(s): 346

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 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 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.

 

 

Author

SYSU

 

 

Source

2014 Multi-University Training Contest 9 

 

 

/**
给出一个数列：a[i]，然后
b[i]：表示在 i 前面的项，如果有a[i]的倍数（要最靠近i的），那么b[i]就等于这个数，如果没有那么b[i] = a[i];
c[i]：表示在 i 后面的项，如果有a[i]的倍数（要最靠近i的），那么c[i] 就等于这个数，如果没有那么c[i] = a[i];
**/
#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
#include<vector>
using namespace std;

int a[100002],b[100002],c[100002];
vector<int>Q[100002];
int hash1[100002];
int main()
{
    int n;
    int MAX,MIN,k;
    for(int i=2;i<=100000;i++)
        for(int j=i;j<=100000;j=j+i)
        Q[i].push_back(j);
    while(scanf("%d",&n)>0)
    {
        if(n==0)break;
        for(int i=1; i<=n; i++){
            scanf("%d",&a[i]);
            b[i] = c[i] = a[i];
        }
        memset(hash1,0,sizeof(hash1));
        hash1[a[1]] = 1;
        for(int i=2;i<=n;i++)
        {
            if(a[i]==1){
                b[i] = a[i-1];
                continue;
            }
           k = Q[a[i]].size();
           MAX = -1;
           for(int j=0;j<k;j++)
            if(hash1[Q[a[i]][j]]!=0 && MAX<hash1[Q[a[i]][j]])
            MAX = hash1[Q[a[i]][j]];

            if(MAX == -1);
            else b[i] = a[MAX];
            hash1[a[i]] = i;
        }
        memset(hash1,0,sizeof(hash1));
        hash1[a[n]] = n;
        for(int i=n-1;i>=1;i--)
        {
            if(a[i]==1) { c[i] = a[i+1]; continue;}
            MIN = 111111111;
            k = Q[a[i]].size();
            for(int j=0;j<k;j++)
                if(hash1[Q[a[i]][j]]!=0 && MIN>hash1[Q[a[i]][j]])
                MIN = hash1[Q[a[i]][j]];
            if(MIN ==111111111 );
            else c[i] = a[MIN];
            hash1[a[i]] = i;
        }
        __int64 sum = 0;
        for(int i=1;i<=n;i++)
            sum = sum+((__int64)b[i])*c[i];
        printf("%I64d\n",sum);
    }
    return 0;
}