HDU4961——Boring Sum(数论)

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本文介绍了一个名为BoringSum的问题，该问题要求计算一个整数序列的“无聊和”。通过定义左侧最近倍数和右侧最近倍数的概念，利用数组记录最近倍数的位置，并通过枚举约数更新记录，最终实现O(n sqrt(n))复杂度下解决问题。

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Boring Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 215 Accepted Submission(s): 103

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：

给出数组a[i]

定义b[i]为 a[i]左侧第一个为a[i]倍数的数字，如果没有，则b[i]=a[i]。

定义c[i]为 a[i]右侧第一个为a[i]倍数的数字，如果没有，则c[i]=a[i]。

求∑(b[i]*c[i])

分析：

数组vis[i] 表示最新的是i的倍数的数字。

从左向右，如果vis[ a[i] ]>0 b[i]=vis[ a[i] ],否则b[i]=a[i]，然后对于每一个a[i]，枚举他的约数并且更新数组vis。

再从右向左计算c[i]。

最后求出答案。

算法复杂度（n*sqrt（n））

#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#define INF 0x7fffffff
using namespace std;

const int N = 1e5 + 10;

int vis[N],a[N],b[N],c[N];

void update(int x)
{
    int k=sqrt(1.0*x);
    for(int i=1;i<=k;i++)
    {
        if(x%i) continue;

        vis[i]=vis[x/i]=x;
    }
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF&&n)
    {
        memset(vis,0,sizeof(vis));
        memset(b,0,sizeof(b));
        memset(c,0,sizeof(c));

        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);

        for(int i=0;i<n;i++)
        {
            if(vis[a[i]]) b[i]=vis[a[i]];
            else b[i]=a[i];
            update(a[i]);
        }

        memset(vis,0,sizeof(vis));

        for(int i=n-1;i>=0;i--)
        {
            if(vis[a[i]]) c[i]=vis[a[i]];
            else c[i]=a[i];
            update(a[i]);
        }

        long long sum=0;
        for(int i=0;i<n;i++)
            sum+=1LL*c[i]*b[i];

        printf("%I64d\n",sum);
    }

    return 0;
}