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HDU4961：Boring Sum

2024-07-18 14:23:03   224人阅读

Problem Description
Number theory is interesting, while this problem is boring.

Here is the problem. Given an integer sequence a1, a2, …, an, let S(i) = {j|1<=j<i, and aj is a multiple of ai}. 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 af(i). Similarly, let T(i) = {j|i<j<=n, and aj is a multiple of ai}. If T(i) is not empty, let g(i) be the minimum integer in T(i); otherwise, g(i) = i. Now we define ci as ag(i). The boring sum of this sequence is defined as b1 * c1 + b2 * c2 + … + bn * cn.

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, …, an (1<= ai<=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.
 

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
#define up(i,x,y) for(i=x;i<=y;i++)
#define down(i,x,y) for(i=x;i>=y;i--)
#define mem(a,b) memset(a,b,sizeof(a))
#define w(x) while(x)
#define l 100005
#define ll __int64
#define s(a) scanf("%I64d",&a)
ll vis[l],a[l],b[l],c[l],n,i,j,ans;

int main()
{
    w((s(n),n))
    {
        up(i,1,n)
        s(a[i]);
        mem(vis,0);
        up(i,1,n)
        {
            if(vis[a[i]])
                b[i]=a[vis[a[i]]];
            else
                b[i]=a[i];
            up(j,1,sqrt(a[i]*1.0)+1)
            {
                if(a[i]%j==0)
                    vis[j]=vis[a[i]/j]=i;
            }
        }
        mem(vis,0);
        down(i,n,1)
        {
            if(vis[a[i]])
                c[i]=a[vis[a[i]]];
            else
                c[i]=a[i];
            up(j,1,sqrt(a[i]*1.0)+1)
            {
                if(a[i]%j==0)
                    vis[j]=vis[a[i]/j]=i;
            }
        }
        ans=0;
        up(i,1,n)
        ans+=b[i]*c[i];
        printf("%I64d\n",ans);
    }

    return 0;
}




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