hdu 4961

Boring Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 155    Accepted Submission(s): 81

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

 

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这题用 筛法标记就好，最最重要的是，vector[]数组要开在 main()函数外面，不然会爆栈！！

代码：

#include <iostream>#include<cstdio>#include<cstring>#include<vector>#include<algorithm>#define M 100000using namespace std;typedef __int64 ll;int a[M+5],mark[M+5],b[M+5],c[M+5];vector<int>adj[M+5];  // 这里，开在 main（）函数里面，会爆栈int main(){    int n;    for(int i=1;i<=M;i++)        {            for(int j=1;j<=M/i;j++)                adj[i*j].push_back(i);  //  求因子        }    while(~scanf("%d",&n) && n)    {        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);        memset(mark,0,sizeof(mark));        for(int i=1;i<=n;i++)        {            if(mark[a[i]]==0) b[i]=a[i];            else b[i]=mark[a[i]];            for(int j=0;j<adj[a[i]].size();j++)                {                    int d=adj[a[i]][j];                    mark[d]=a[i];                }        }    //    for(int i=1;i<=n;i++)      //      printf("%d : %d \n",i,b[i]);        memset(mark,0,sizeof(mark));        for(int i=n;i>=1;i--)        {            if(mark[a[i]]==0) c[i]=a[i];            else c[i]=mark[a[i]];            for(int j=0;j<adj[a[i]].size();j++)                {                    int d=adj[a[i]][j];                    mark[d]=a[i];                }        }      //  for(int i=1;i<=n;i++)        //    printf("%d :%d \n",i,c[i]);        ll ans=0;        for(int i=1;i<=n;i++)            ans+=(ll)b[i]*c[i];        printf("%I64d\n",ans);    }    return 0;}