hdu 4961 枚举约数O(n*sqrt(n))

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本文介绍了一个名为BoringSum的问题，该问题涉及整数序列及其倍数关系，并定义了一种特殊的和，即“无聊和”。文章详细解释了如何通过算法找到序列中每个元素左侧和右侧最近的倍数下标，并给出了完整的实现代码。

Boring Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1099 Accepted Submission(s): 517

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

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

Recommend

f[i]是a[i]左边最近的倍数的下标，g[i]是a[i]右边最近的倍数的下标。

用h[k]，记录k的倍数上次出现的位置。求f[i]时，从左往右扫，扫到a[i]时，f[i]=h[a[i]],并且枚举出a[i]所有约数j,更新h[j]=h[a[i]/j]=i.因为枚举约数复杂度为O(sqrt(n))，对于10^5的数据，O(n*sqrt(n))能在规定时间内完成。

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

#define maxn 100001
int f[maxn],g[maxn],a[maxn];
int h[maxn];

int main()
{
    int n;
    while(~scanf("%d", &n) && n){
        for(int i = 1; i <= n; i++)
            scanf("%d", a+i);

        memset(h, 0, sizeof(h));

        for(int i = 1; i <= n; i++){
            f[i] = h[a[i]]? h[a[i]]:i;

            for(int j = 1; j*j <= a[i]; j++)
            if(a[i]%j == 0){
                h[a[i]/j] = h[j] = i;
            }
        }

        memset(h,0,sizeof(h));
        for(int i = n; i > 0; i--){
            g[i] = h[a[i]]?h[a[i]]:i;

            for(int j = 1; j*j <= a[i]; j++)
            if(a[i]%j == 0){
                h[a[i]/j]=h[j] = i;
            }
        }

        long long ans = 0;
        for(int i =1; i <= n; i++)
            ans += (long long)a[f[i]]*a[g[i]];
        printf("%I64d\n", ans);
    }
    return 0;
}