hdu4961 Boring Sum(数学)

题目：

Boring Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1213    Accepted Submission(s): 569

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.
 

 

Author

SYSU

 

Source

2014 Multi-University Training Contest 9

 

题意：给一个数组，定义f[i]是左边最靠近a[i]并且是a[i]倍数的值的下标，g[i]右边最靠近a[i]并且是a[i]倍数的值的下标，b[i]=a[f[i]],c[i]=a[g[i]]，求b1*c1+b2*c2+...+bn*cn.

思路：其实就是怎么快速找出最靠近一个数并且是它倍数的数。我们可以先暴力枚举出每个数的约数（用类似素数筛法的方法，nlogn），然后用一个数组来维护一个约数最后出现的位置。先从左往右枚举数组中的每一个元素，如果它已经出现在了约数数组里，那么对应的约数数组的值就是它的左边最靠近它的倍数的下标。然后将它的左右约数填进约数数组里。再从右往左枚举数组就得到了它的右边最靠近它的倍数的下标。

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<vector>using namespace std;const int MAXN=100000;int a[MAXN+5];vector<int>f[MAXN+5];int F[MAXN+5],G[MAXN+5];int b[MAXN+5],c[MAXN+5];void init(){for(int i=1;i<=MAXN;i++)f[i].clear();for(int i=1;i<=MAXN;i++)for(int j=i;j<=MAXN;j+=i)f[j].push_back(i);}int vis[MAXN+5];int main(){int n;init();while(scanf("%d",&n)!=EOF){if(!n)break;for(int i=1;i<=n;i++){scanf("%d",&a[i]);}memset(vis,0,sizeof vis);for(int i=1;i<=n;i++){if(!vis[a[i]]){vis[a[i]]=i;F[i]=i;}else{F[i]=vis[a[i]];vis[a[i]]=i;}for(int j=0;j<f[a[i]].size();j++){vis[f[a[i]][j]]=i;}}//printf("%d ",F[i]);//printf("\n");memset(vis,0,sizeof(vis));for(int i=n;i>=1;i--){if(!vis[a[i]]){vis[a[i]]=i;G[i]=i;}else{G[i]=vis[a[i]];}for(int j=0;j<f[a[i]].size();j++){vis[f[a[i]][j]]=i;}}__int64 ans=0;for(int i=1;i<=n;i++){b[i]=a[F[i]];c[i]=a[G[i]];ans=ans+(__int64)b[i]*c[i];}printf("%I64d\n",ans);}return 0;}