HDU 4961 Boring Sum 打表、更新

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點擊開啟連結Boring SumTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 715    Accepted Submission(s): 351


Problem DescriptionNumber 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. 
InputThe 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. 
OutputOutput the answer in a line. 
Sample Input

51 4 2 3 90

 
Sample Output

136HintIn 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. 

 
AuthorSYSU 
Source2014 Multi-University Training Contest 9 對於輸入的數列，從前往後掃一遍，對於每個數都要更新為距離它左邊最近的倍數的值，如果沒有則為次數。同樣從後往前掃一遍，對於每個數都要更新為距離它右邊最近的倍數的值，如果沒有也為次數。將所有數的約數打個表存起來，然後掃兩遍分別記錄b[]和c[]，掃的過程中要隨時更新。

//203MS8904K#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];          //因為打表的時候沒有算上約數1，所以要加上             for(int j=0;j<v[a[i]].size();j++)//更新a[i]的約數                 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;}