Boring Sum（hdu4961）hash

Boring Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 814 Accepted Submission(s): 390

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.

 

 

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>
#define ll __int64
#define maxn 100000+5
#define mem(x) memset(x,0,sizeof(x))


ll a[maxn],b[maxn],c[maxn],sum,n;
ll vis[maxn];//a[i]的下标i

int main()
{
    ll i,j,k,temp;
    while(scanf("%I64d",&n),n)
    {
        mem(b);
        mem(c);
        mem(vis);
        for(i=1;i<=n;i++)
            scanf("%I64d",&a[i]);
        vis[a[1]]=1;
        for(i=2;i<=n;i++)
        {
            for(j=1;j*j<=a[i];j++)
            {
                if(a[i]%j!=0) continue;//取质因子

//                printf("i=%I64d  j=%I64d  a[i]=%I64d  ",i,j,a[i]);

                if(vis[j]!=0)//
                {
                    b[vis[j]]=a[i];
//                    printf("b[vis[j]]=%I64d  vis[j]=%I64d  ",b[vis[j]],vis[j]);
                    vis[j]=0;
                }
                temp=a[i]/j;
 //               printf("temp=%I64d  ",temp);
                if(vis[temp]!=0)//更新
                {
                    b[vis[temp]]=a[i];
  //                  printf("b[vis[temp]]=%I64d  vis[temp]=%I64d",b[vis[temp]],vis[temp]);
                    vis[temp]=0;
                }
   //             printf("
");
            }
            vis[a[i]]=i;
        }
        for(i=1;i<=n;i++)
            if(b[i]==0)
                b[i]=a[i];
        mem(vis);
        vis[a[n]]=n;

        for(i=n-1;i>=1;i--)
        {
            for(j=1;j*j<=a[i];j++)
            {
                if(a[i]%j!=0) continue;//取质因子
                if(vis[j]!=0)//
                {
                    c[vis[j]]=a[i];
                    vis[j]=0;
                }
                temp=a[i]/j;
                if(vis[temp]!=0)//更新
                {
                    c[vis[temp]]=a[i];
                    vis[temp]=0;
                }
            }
            vis[a[i]]=i;
        }
        for(i=1;i<=n;i++)
            if(c[i]==0)
                c[i]=a[i];
  //      for(i=1;i<=n;i++)
  //      printf("b:%I64d	c:%I64d
",b[i],c[i]);
        sum=0;
        for(i=1;i<=n;i++)
            sum+=b[i]*c[i];
        printf("%I64d
",sum);
    }
    return 0;
}