HDU-4961 Boring Sum （模拟）

Boring Sum

http://acm.hdu.edu.cn/showproblem.php?pid=4961

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

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.

又是无限接近正解而没有写出来，原本想着维护一个[1,i)中是a[i]点倍数点下标点最值，但是两个最值都要维护，就不会了。。。

先正着扫一遍，维护[1,i)中是a[i]点倍数点下标点最大值，并及时更新b[i]；再倒着扫一遍，维护(i,n]中是a[i]点倍数点下标点最小值，并及时更新c[i]

最后相乘得出答案

#include <cstdio>#include <cstring>#include <vector>using namespace std;const int MAXN=100005;int n,f,g;int a[MAXN],b[MAXN],c[MAXN],indx[MAXN];long long ans;vector<int> factor[MAXN];//factor[i]表示i点因子的表int main() {    for(int j=1;j<=MAXN;++j)        for(int i=j;i<=MAXN;i+=j)            factor[i].push_back(j);    while(scanf("%d",&n),n!=0) {        memset(indx,0,sizeof(indx));        for(int i=1;i<=n;++i) {//indx[x]表示[1,i)中是a[i]的倍数的最大下标            scanf("%d",a+i);            f=indx[a[i]]==0?i:indx[a[i]];            b[i]=a[f];            for(int j=0;j<factor[a[i]].size();++j)                indx[factor[a[i]][j]]=max(indx[factor[a[i]][j]],i);        }        memset(indx,0x3f,sizeof(indx));        for(int i=n;i>0;--i) {//indx[x]表示(i,n]中是a[i]的倍数的最小下标            g=indx[a[i]]==0x3f3f3f3f?i:indx[a[i]];            c[i]=a[g];            for(int j=0;j<factor[a[i]].size();++j)                indx[factor[a[i]][j]]=min(indx[factor[a[i]][j]],i);        }        ans=0;        for(int i=1;i<=n;++i)            ans+=(long long)b[i]*c[i];        printf("%I64d\n",ans);    }    return 0;}