HDU4961Boring Sum(在线更新方法)

Boring Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 446    Accepted Submission(s): 227

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. 
    
    
    
    
     
   
   
   
   

Author

SYSU

Source

2014 Multi-University Training Contest 9

 题意:给你一个数组，让你生成两个新的数组，A要求每个数如果能在它的前面找个最近的一个是它倍数的数，那就变成那个数，否则是自己，C是往后找，最后每对应的b,c相乘之和。

#include<stdio.h>
#include<string.h>

__int64 fact[100005][129],k[100005];
void setprim()
{
    memset(k,0,sizeof(k));
    for(__int64 i=1;i<=100000;i++)
    {
        for(__int64 j=i;j<=100000;j+=i)
        {
            fact[j][++k[j]]=i;
        }
    }
}
 __int64 a[100005],b[100005],c[100005],index[100005];
int main()
{
    setprim();
    __int64 i,j,sum,n;
    while(scanf("%I64d",&n)>0&&n)
    {
        for( i=1;i<=n;i++)
            scanf("%I64d",&a[i]);
        memset(index,0,sizeof(index));
        for(i=1;i<=n;i++)
        {
            if(index[a[i]])b[i]=a[index[a[i]]];
            else b[i]=a[i];
            for(j=1;j<=k[a[i]];j++)
            index[fact[a[i]][j]]=i;
        }
        sum=0;
        memset(index,0,sizeof(index));
        for(i=n;i>=1;i--)
        {
            if(index[a[i]])c[i]=a[index[a[i]]];
            else c[i]=a[i];
            for(j=1;j<=k[a[i]];j++)
            index[fact[a[i]][j]]=i;
        }
        for( i=1;i<=n;i++)
        sum+=b[i]*c[i];
        printf("%I64d\n",sum);
    }
}