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hdu 4961 Boring Sum（数学题）

2024-07-18 14:22:59   227人阅读

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


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

题意：

给出一个数列：a[i]，然后

b[i]：表示在 i 前面的项，如果有a[i]的倍数（要最靠近i的），那么b[i]就等于这个数，如果没有那么b[i] = a[i];

c[i]：表示在 i 后面的项，如果有a[i]的倍数（要最靠近i的），那么c[i] 就等于这个数，如果没有那么c[i] = a[i];

思路：

//先打表，把每个数的约数存在vector里；
//然后从前往后扫一遍，结果存在b[i]，

//Ps：如果不清楚为什么从前往后扫一遍就是最靠近的那个数可调试一下（案例：9 6 3 2 1）；
//然后从后往前扫一遍，结果存在c[i]，

//Ps：如果不清楚为什么从后往前扫一遍就是最靠近的那个数可调试一下（案例：1 2 3 6 9）；
//最后计算b[i]*c[i]的和即可。


代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#define maxn 100000+17
using namespace std;
typedef __int64 LL;
int vis[maxn];
int a[maxn], b[maxn], c[maxn];
vector<int>V[maxn];
void init()
{
    for(int i = 0; i < maxn; i++)
        V[i].clear();
    for(int i = 1; i <= maxn; i++)
    {
        for(int j = 1; j*i <= maxn; j++)//每个数相应的约数
        {
            V[i*j].push_back(i);//i是哪些数的约数
        }
    }
}
int main()
{
    int n;
    init();
    while(scanf("%d",&n) && n)
    {
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&a[i]);
        }
        memset(vis,0,sizeof(vis));
        for(int i = 1; i <= n; i++)
        {
            if(vis[a[i]] == 0)
                b[i]=a[i];
            else
                b[i]=vis[a[i]];//a[i]的倍数
            for(int j = 0; j < V[a[i]].size(); j++)
                vis[V[a[i]][j]] = a[i];//V[a[i]][j]为a[i]的约数
        }
        memset(vis,0,sizeof(vis));
        for(int i = n; i >= 1; i--)
        {
            if(vis[a[i]] == 0)
                c[i] = a[i];
            else
                c[i] = vis[a[i]];
            for(int j = 0; j < V[a[i]].size(); j++)
                vis[V[a[i]][j]] = a[i];
        }
        LL sum=0;
        for(int i = 1; i <= n; i++)
        {
            sum += (LL)b[i]*(LL)c[i];
        }
        printf("%I64d\n",sum);
    }
    return 0;
}



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