hdu 4961 数论 o(nlogn)

Boring Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 60    Accepted Submission(s): 30

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.

 

Source

2014 Multi-University Training Contest 9

 

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题解：对于输入的数列，从前往后扫描一遍，对于每个数，都更新一下它的约数的左边最近倍数的值（b值）；

同样地，从后往前扫描一遍，对于每个数，都更新一下它的约数的右边最近倍数的值（c值）。最后直接求所有b*c的和即可。

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>

#define N 100005
#define M 15
#define mod 1000000007
#define mod2 100000000
#define ll long long
#define maxi(a,b) (a)>(b)? (a) : (b)
#define mini(a,b) (a)<(b)? (a) : (b)

using namespace std;

int n;
ll a[N],b[N],c[N];
int vis[N];
ll ans;

int main()
{
    int i;
   // freopen("data.in","r",stdin);
    //scanf("%d",&T);
    //for(int cnt=1;cnt<=T;cnt++)
    //while(T--)
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0) break;
        ans=0;
        memset(b,0,sizeof(b));
        memset(c,0,sizeof(c));
        memset(vis,0,sizeof(vis));
        for(i=1;i<=n;i++){
            scanf("%I64d",&a[i]);
        }

        vis[ a[1] ]=1;
        for(i=2;i<=n;i++){
            for(ll j=1;j*j<=a[i];j++){
                if(a[i]%j!=0) continue;
                if(vis[j]!=0){
                    b[ vis[j] ]=a[i];
                    vis[j]=0;
                }
                ll te=a[i]/j;
                if(vis[te]!=0){
                    b[ vis[te] ]=a[i];
                    vis[te]=0;
                }
            }
            vis[ a[i] ]=i;
        }


        for(i=1;i<=n;i++){
            if(b[i]==0) b[i]=a[i];
        }

        memset(vis,0,sizeof(vis));
        vis[ a[n] ]=n;
        for(i=n-1;i>=1;i--){
            for(ll j=1;j*j<=a[i];j++){
                if(a[i]%j!=0) continue;
                if(vis[j]!=0){
                    c[ vis[j] ]=a[i];
                    vis[j]=0;
                }
                ll te=a[i]/j;
                if(vis[te]!=0){
                    c[ vis[te] ]=a[i];
                    vis[te]=0;
                }
            }
            vis[ a[i] ]=i;
        }

        for(i=1;i<=n;i++){
            if(c[i]==0) c[i]=a[i];
        }

        for(i=1;i<=n;i++){
            ans+=b[i]*c[i];
        }
        printf("%I64d
",ans);

    }

    return 0;
}