HDU 4961 Boring Sum

Boring Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 739    Accepted Submission(s): 363

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

51 4 2 3 90

 

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.
 

 

Source

2014 Multi-University Training Contest 9


题目链接  :http://acm.hdu.edu.cn/showproblem.php?pid=4961


题目大意  :输入一个数列a[i]，数列b[j]的组成为：如果在a[i]的前i-1项中有a[i]的倍数，取最接近i的a[i]值为b[i]值，如果没有那么b[i] = a[i]

数列c[i]的组成为：如果在a[i]的i+1到n项中有a[i]的倍数，取最接近i的a[i]值为c[i]值，如果没有那么c[i] = a[i]

题目分析 :左右扫一下算出b,c数组

#include <cstdio>#include <cstring>#define ll long longint const MAX = 1e5 + 10;int n, vis[MAX];ll a[MAX], b[MAX], c[MAX], ans;int main(){    while(scanf("%d", &n) != EOF && n)    {        ans = 0;        memset(b, 0, sizeof(b));        memset(c, 0, sizeof(c));        memset(vis, 0, sizeof(vis));        for(int i = 1; i <= n; i++)            scanf("%I64d", &a[i]);        vis[ a[1] ] = 1;        for(int i = 2; i <= n; i++)        {            for(ll j = 1; j * j <= a[i]; j++)            {                if(a[i] % j)                     continue;                if(vis[j])                {                    b[ vis[j] ] = a[i];                    vis[j] = 0;                }                ll tmp = a[i] / j;                if(vis[tmp])                {                    b[ vis[tmp] ] = a[i];                    vis[tmp] = 0;                }            }            vis[ a[i] ] = i;        }        for(int i = 1; i <= n; i++)            if(b[i] == 0)                 b[i] = a[i];        memset(vis, 0, sizeof(vis));        vis[ a[n] ] = n;        for(int i = n - 1; i >= 1; i--)        {            for(ll j = 1; j * j <= a[i]; j++)            {                if(a[i] % j)                     continue;                if(vis[j])                {                    c[ vis[j] ] = a[i];                    vis[j] = 0;                }                ll tmp = a[i] / j;                if(vis[tmp])                {                    c[ vis[tmp] ] = a[i];                    vis[tmp] = 0;                }            }            vis[ a[i] ] = i;        }        for(int i = 1; i <= n; i++)            if(c[i] == 0)                 c[i] = a[i];        for(int i = 1; i <= n; i++)            ans += b[i] * c[i];        printf("%I64d\n",ans);    }}