2014多校联合第9场1002||hdu 4691 质因子

http://acm.hdu.edu.cn/showproblem.php?pid=4961

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. 
    
    
    
    
     
   
   
   
   

题意： 

         给出n个数的数列a，bi的取值为在1 <= j < i之间如果存在aj % ai == 0，则取最大下标的值赋给bi，如果不存在，则bi = ai；ci的取值为在i < j <= n之间如果存在aj % ai == 0，则取最小下标值赋给bi，如果不存在，则ci = ai。求b1 * c1 + b2 * c2 + ... + bn * cn的和。

思路：

          如果直接暴力的话一定会超时，所以我们可以开一个vis数组来记录每一个值所对应的最大的下标是多少。即每查找ai，分解出ai的质因子，更新vis数组。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <cmath>
using namespace std;
typedef __int64 LL;
int a[100005],vis[100005],b[100005],c[100005],n;
int main()
{
    while(~scanf("%d",&n))
    {
        if(n==0)
            break;
        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]])
                b[i]=a[vis[a[i]]];
            else
                b[i]=a[i];
            for(int j=1; j<=(int)sqrt((double)a[i]) + 1; j++)
            {
                if(a[i]%j==0)
                {
                    vis[a[i]/j]=i;
                    vis[j]=i;
                }
            }
        }
        memset(vis,0,sizeof(vis));
        for(int i=n; i>=1; i--)
        {
            if(vis[a[i]])
                c[i]=a[vis[a[i]]];
            else
                c[i]=a[i];
            for(int j=1; j<=(int)sqrt((double)a[i])+1; j++)
            {
                if(a[i]%j==0)
                {
                    vis[a[i]/j]=i;
                    vis[j]=i;
                }
            }
        }
       /*
       
        for(int i=1;i<=n;i++)
            printf("%d ",b[i]);
        printf("\n");
        for(int j=1;j<=n;j++)
            printf("%d ",c[j]);
        printf("\n");
        
        */
        LL ans=0;
        for(int i=1; i<=n; i++)
            ans+=(LL)b[i]*c[i];
        printf("%I64d\n",ans);
    }
    return 0;
}