HDU 4961 Boring Sum

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本文介绍了一个名为BoringSum的问题，该问题涉及整数序列处理、最大最小值查找及复杂求和运算。通过定义特定的函数f(i)和g(i)，实现对序列的特殊求和。文章提供了一个C++实现方案，包括输入处理、关键算法实现及输出结果。

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Boring Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 114 Accepted Submission(s): 59

Problem Description

Number theory is interesting, while this problem is boring.

Here is the problem. Given an integer sequence a ₁, a ₂, …, 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 b ₁ * c ₁ + b ₂ * c ₂ + … + 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 a ₁, a ₂, …, a _n (1<= a _i<=100000).

The input is terminated by n = 0.

Output

Output the answer in a line.

Sample Input

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

两数相乘会爆int WA了一发。。类型转换下就好了。。

#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>

using namespace std;

#define PB push_back
#define MP make_pair
#define CLR(vis) memset(vis,0,sizeof(vis))
#define MST(vis,pos) memset(vis,pos,sizeof(vis))
#define MAX3(a,b,c) max(a,max(b,c))
#define MAX4(a,b,c,d) max(max(a,b),max(c,d))
#define MIN3(a,b,c) min(a,min(b,c))
#define MIN4(a,b,c,d) min(min(a,b),min(c,d))
#define PI acos(-1.0)
#define INF 0x7FFFFFFF
#define LINF 1000000000000000000LL
#define eps 1e-8

typedef long long ll;
typedef unsigned long long ull;


const int maxn=100000+100;
int a[maxn];
int f[maxn];
int g[maxn];

struct node{
    int vis;
    vector<int> id;
}e[maxn];


int main()
{
    int n;
    while(cin>>n && n)
    {
        CLR(e);
        int max_val=-1;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            max_val=max(max_val,a[i]);
            if(e[a[i]].vis==0) e[a[i]].id.clear();
            e[a[i]].vis=1;
            e[a[i]].id.PB(i);
        }
        int min_id,max_id;
        for(int i=1;i<=n;i++)
        {
            min_id=INF,max_id=-1;
            int ans=a[i];
            while(ans<=max_val)
            {
                if(e[ans].vis==1)
                {
                    for(int j=0;j<e[ans].id.size();j++)
                    {
                        int id=e[ans].id[j];
                        if(id<i && id>=1)
                           max_id=max(max_id,id);
                        if(id>i && id<=n)
                           min_id=min(min_id,id);
                    }
                }
                ans+=a[i];
            }
            if(max_id!=-1)   f[i]=max_id;
            else             f[i]=i;
            if(min_id!=INF)  g[i]=min_id;
            else             g[i]=i;
        }
        ll sum=0;
        for(int i=1;i<=n;i++)
            sum+=(ll)a[f[i]]*a[g[i]];
        cout<<sum<<endl;


    }
    return 0;
}