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HDU 4961 Boring Sum 打表、更新

2024-07-18 17:14:21   226人阅读

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Boring Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 715    Accepted Submission(s): 351


Problem Description
Number theory is interesting, while this problem is boring.

Here is the problem. Given an integer sequence a1, a2, …, an, let S(i) = {j|1<=j<i, and aj is a multiple of ai}. 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 af(i). Similarly, let T(i) = {j|i<j<=n, and aj is a multiple of ai}. If T(i) is not empty, let g(i) be the minimum integer in T(i); otherwise, g(i) = i. Now we define ci as ag(i). The boring sum of this sequence is defined as b1 * c1 + b2 * c2 + … + bn * cn.

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, …, an (1<= ai<=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.
 

Author
SYSU
 

Source
2014 Multi-University Training Contest 9
 
对于输入的数列，从前往后扫一遍，对于每个数都要更新为距离它左边最近的倍数的值，如果没有则为次数。同样从后往前扫一遍，对于每个数都要更新为距离它右边最近的倍数的值，如果没有也为次数。
将所有数的约数打个表存起来，然后扫两遍分别记录b[]和c[]，扫的过程中要随时更新。
//203MS	8904K
#include<stdio.h>
#include<string.h>
#include<vector>
using namespace std;
vector<int>v[100007];
int a[100007],b[100007],c[100007],vis[100007];
int main()
{
    int n;
    for(int i=2;i<100007;i++)
    {
        for(int j=i;j<100007;j+=i)
            v[j].push_back(i);
    }
    while(scanf("%d",&n),n)
    {
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)
        {
             if(!vis[a[i]])b[i]=a[i];
             else b[i]=vis[a[i]];
             vis[1]=a[i];          //因为打表的时候没有算上约数1，所以要加上
             for(int j=0;j<v[a[i]].size();j++)//更新a[i]的约数
                 vis[v[a[i]][j]]=a[i];
        }
        memset(vis,0,sizeof(vis));
        for(int i=n;i>=1;i--)
        {
             if(!vis[a[i]])c[i]=a[i];
             else c[i]=vis[a[i]];
             vis[1]=a[i];
             for(int j=0;j<v[a[i]].size();j++)
                 vis[v[a[i]][j]]=a[i];
        }
        __int64 ans=0;
        for(int i=1;i<=n;i++)
            ans+=(__int64)b[i]*(__int64)c[i];
        printf("%I64d\n",ans);
    }
    return 0;
}



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