HDU 5288 OO’s Sequence (二分查找)_sequence二分-优快云博客

本文链接：https://blog.youkuaiyun.com/Fungyow/article/details/46992575

本文介绍了一道算法竞赛题目“OO’sSequence”的解题思路与实现细节。该题要求计算特定条件下数列中不可整除其他元素的数量总和。通过定义辅助数组l[i],r[i]并使用二分查找技巧求解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

OO’s Sequence

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 569 Accepted Submission(s): 205

Problem Description

OO has got a array A of size n ,defined a function f(l,r) represent the number of i (l<=i<=r) , that there's no j(l<=j<=r,j<>i) satisfy a_i mod a_j=0,now OO want to know

\sum i = 1 n \sum j = i n f (i, j) m o d （ 109 + 7 ） .

Input

There are multiple test cases. Please process till EOF.
In each test case:
First line: an integer n(n<=10^5) indicating the size of array
Second line:contain n numbers a_i(0<a_i<=10000)

Output

For each tests: ouput a line contain a number ans.

Sample Input

5
1 2 3 4 5

Sample Output

23


题意：求对于所有i<=j，数列a[i],a[i+1]...a[j]中不能整除其他所有元素的元素数量的总和。
思路：2015多校赛第一场，比赛中想不出，多校赛的简单题对我等渣渣实在不简单。过后看题解，定义两个数组l[i], r[i],表示第i个数左侧和右侧最接近它且值是a[i]因子的数字的位置。第i个数只能在l[i]+1到r[i]-1区间的包含a[i]的子区间有贡献
，除了这些区间a[i]都因其因子的存在而没有贡献。贡献值为（i-l[i]）*(r[i]-i)。因此问题转为求l[i],r[i]。具体如代码所示，过程中用到二分查找。

#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <string.h>
#include <vector>
#define MOD 1000000007
using namespace std;

int l[100009], r[100009], a[100009];
vector<int> w[10009];
int findr(int k, int n, int i)
{
    int lb=0, ub=n-1;
    while(ub-lb>1)
    {
        int mid=(ub+lb)/2;
        if(w[i][mid] > k)
        {
            ub=mid;
        }
        else
        {
            lb=mid;
        }
    }
    return w[i][ub];
}
int findl(int k, int n, int i)
{
    int lb=0, ub=n+1;
    while(ub-lb>1)
    {
        int mid=(ub+lb)/2;
        if(w[i][mid] < k)
        {
            lb=mid;
        }
        else
        {
            ub=mid;
        }
    }
    return w[i][lb];
}
int main()
{
    int n;
    while(~scanf("%d", &n))
    {
        for(int i=0; i<10009; i++)
        {
            w[i].clear();
            w[i].push_back(0);
        }
        for(int i=1; i<=n; i++)
        {
            scanf("%d", &a[i]);
            l[i]=0, r[i]=n+1;
            w[a[i]].push_back(i);
        }
        for(int i=0; i<10009; i++)
        {
            w[i].push_back(n+1);
        }
        long long ans=0;
        for(int i=1; i<=n; i++)
        {

            for(int j=1; j<=sqrt(a[i]); j++)
            {

                if(a[i]%j != 0)
                    continue;
                else
                {
                    int c=a[i]/j;
                    l[i]=max(l[i], findl(i, w[c].size(), c));
                    r[i]=min(r[i], findr(i, w[c].size(), c));
                    c=j;
                    l[i]=max(l[i], findl(i, w[c].size(), c));
                    r[i]=min(r[i], findr(i, w[c].size(), c));
                }
            }
            ans+=(long long)(i-l[i])*(r[i]-i);
        }
        printf("%lld\n", ans%MOD);
    }
    return 0;
}