HDU4961——Boring Sum(數論)

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.

題意：

給出數組a[i]

定義b[i]爲 a[i]左側第一個爲a[i]倍數的數字，如果沒有，則b[i]=a[i]。

定義c[i]爲 a[i]右側第一個爲a[i]倍數的數字，如果沒有，則c[i]=a[i]。

求∑(b[i]*c[i]) 

分析：

數組vis[i] 表示最新的是i的倍數的數字。

從左向右，如果vis[ a[i] ]>0 b[i]=vis[ a[i] ],否則b[i]=a[i]，然後對於每一個a[i]，枚舉他的約數並且更新數組vis。

再從右向左計算c[i]。

最後求出答案。

算法複雜度（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;
}